s.n. contents p ge therma ysics

TH

ER

MA

L P

HY

SIC

S

NEET SYLLABUSThermal equilibrium and definition of temperature (zeroth law of Thermodynamics). Heat, temperature, thermal expansion; thermalexpansion of solids, liquids, and gases. Anomalous expansion. Specific heat capacity: Cp, Cv- calorimetry; change of state - latent heat.Heat transfer- conduction and thermal conductivity, convection and radiation. Qualitative ideas of Black Body Radiation, Wein's displacementlaw Newton's law of cooling and Stefan's law. Kinetic theory of gases: Assumptions, concept of pressure. Kinetic energy and temperature;degrees of freedom, law of equipartition of energy (statement only) and application to specific heat capacities of gases; concept of meanfree path. Heat, work and internal energy. First law of thermodynamics. Isothermal and adiabatic processes. Second law of thethermodynamics: Reversible and irreversible processes. Heat engines and refrigerators. Equation of state of a perfect gas, work done oncompressing a gas.

E

CONTENTS S.No. Page

Temperature and Thermal Expansion 1. Zeroth law of thermodynamics 1 2. Scales of temperature 1 3. Thermal expansion 3

Heat 4. Change of state 10 5. Phase diagram 11 6. Law of mixtures; Principle of calorimeter 13

Modes of Heat Transfer 7. Thermal conduction 16 8. Convection 19 9. Thermal radiation 23 10. Stefan's law 26 11. Newton's law of cooling 27 12. Wein's displacement law 29

Kinetic Theory of Gases 13. Ideal gas concept 34 14. Gas laws 35 15. Different speeds of gas molecules 39 16. Expression for pressure of an ideal gas 40 17. Degree of freedom and law of equipartition 41

of energy 18. Different K.E. of gas 42 19. Mean free path 43

Thermodynamics 20. First law of thermodynamics 50 21. Different processes 53 22. Relation between degrees of freedom and specific 57

heat of gas 23. Heat engine and refrigerator 64 24. Second law of thermodynamics 65 25. Reversible and irreversible process 65 26. Carnot cycle 66 27. Exercise -I (Conceptual Questions) 73 28. Exercise-II (Previous Years Questions) 98 29. Exercise-III (Analytical Questions) 109 30. Exercise-IV (Assertion & Reason) 113

RUDOLF CLAUSIUS (1822-1888)

Rudolf Clausius, born in Poland, is generally regarded as the

discoverer of the Second Law of Thermodynamics. Based on the

work of Carnot and Thomson, Clausius arrived at the important

notion of entropy that led him to a fundamental version of the

Second Law of Thermodynamics that states that the entropy of

an isolated system can never decrease. Clausius also worked on

the kinetic theory of gases and obtained the first reliable estimates

of molecular size, speed, mean free path, etc

LUDWIG BOLTZMANN (1844 – 1906)

Ludwig Boltzmann, born in Vienna, Austria, worked on the kinetic

theory of gases independently of Maxwell. A firm advocate of

atomism, that is basic to kinetic theory, Boltzmann provided a

statistical interpretation of the Second Law of thermodynamics

and the concept of entropy. He is regarded as one of the founders

of classical statistical mechanics. The proportionality constant

connecting energy and temperature in kinetic theory is known

as Boltzmann’s constant in his honour.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

1E

Pre-Medical : PhysicsALLENTEMPERATURE & THERMAL EXPANSION

1. TEMPERATURE & TEMPERATURE SCALES

l Temperature

Temperature may be defined as the degree of hotness or coldness of a body. Heat energy flows from

a body at higher temperature to that at lower temperature until their temperatures become equal. At this

stage, the bodies are said to be in thermal equilibrium.

l Thermal equilibrium

Thermal equilibrium is a situation in which two objects would not exchange energy by heat or electromagnetic

radiation if they were placed in thermal contact. Heat is the transfer of energy from one object to another

object as a result of a difference in temperature between them.

l Zeroth law of thermodynamics

If objects A and B are separately in thermal equilibrium with a third object C (say thermometer), then objects

A and B are in thermal equilibrium with each other. Zeroth law of thermodynamics introduces the concept

of temperature. Two objects (or systems) are said to be in thermal equilibrium if their temperatures are same.

C

A B

In measuring the temperature of a body, it is important that the thermometer should be in thermal equilibrium

with the body whose temperature is to be measured.

l Measurement of Temperature

The branch of thermodynamics which deals with the measurement of temperature is called thermometry. A

thermometer is a device used to measure the temperature of a body. The substances like liquids and gases which

are used in the thermometer are called thermometric substances.

l Different Scales of Temperature

A thermometer can be graduated into following scales :

(a) The Centigrade or Celsius scale (°C)

(b) The Fahrenheit scale (°F)

(c) Kelvin scale (K)

l Comparison between Different Temperature Scales

The general formula for the conversion between different temperature scales is:

K - 273 C F -32 X – LFP= = =100 100 180 UFP – LFP

Where X® Reading in unknown temperature scale, LFP ® Lower Fixed Point, UFP ® Upper Fixed Point

Change in temperature K C F X

100 100 180 UFP LFPD D D D

= = =-

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

2 E

Pre-Medical : Physics ALLENGOLDEN KEY POINTS

l Although the temperature of a body can be raised without limit, it can not be lowered without limit and theoretically

limiting low temperature is taken to be zero of the Kelvin scale (i.e. no negative temperature on Kelvin scaleis possible).

l Though when universe was created 1010 years ago, its temp. was about 1039 K which at present is about3K. The highest laboratory temperature is about 108 K (in fusion test reactor) while lowest 10–10 K (achievedin 1999 through nuclear spin cooling) Theory has established that zero Kelvin temperature can never be achieved

practically.

IllustrationsIllustration 1.

Temperature of a patient is 40°C . Find the temperature on Fahrenheit scale ?Solution :

-F 32180

= -40 0

100Þ F = 104°F

Illustration 2.At what temperature is the Fahrenheit scale reading equal to twice of Celsius ?

Solution :

-F 32180

= -C 0

100 Þ

-2x 32180

= -x 0

100 Þ x = 160

Illustration 3.The lower and upper fixed points of a faulty thermometer are 5 W and 105 W. If the thermometer reads25 W, what is the actual temperature in Celsius scale ?

Solution :-

=25 5 C – 0100 100

Þ C = 20°C

Illustration 4.A thermometer with an arbitrary scale has the ice point at –20° and the steam point at 180°. When the

thermometer reads 5°, a Centigrade thermometer will read(1) 7.5 °C (2) 12.5 °C (3) 16.5 °C (4) –9.37 °C

Solution :

C – 0 t ( 20)100 – 0 180 ( 20)

- -=

- - (Here t = 5°)

Þ+

=C 5 20

100 200 Þ C = 12.5 °C

Illustration 5.The temperature of an iron piece is raised from 30°C to 90°C. What is the change in its temperature onthe Fahrenheit scale and on the Kelvin scale?

SolutionDC=90°–30° = 60°C

Temperature difference on Fahrenheit Scale ( )9 9F C 60 C 108 F

5 5D = D = ° = °

Temperature difference on Kelvin Scale K C 60KD = D =

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

3E

Pre-Medical : PhysicsALLEN2. THERMAL EXPANSION

When matter is heated without any change in its state, it usually expands. According to atomic theory ofmatter, asymmetry in potential energy curve is responsible for thermal expansion. As with rise in temperaturethe amplitude of vibration increases and hence energy of atoms increases, hence the average distance betweenthe atom increases. So the matter as a whole expands.

E2

E1

rr0

r1 r2

Potential energy

Thermal expansion arises because the curve is not symmetrical about the equilibrium position r.

As the temperature rises the energy of the atom increases.The mean position when the energy is E is not

the same as that when the energy is E .

0

2

1

l Thermal expansion is minimum in case of solids but maximum in case of gases because intermolecular forcesare maximum in solids but minimum in gases.

l Solids can expand in one dimension (Linear expansion), two dimensions (Superficial expansion) and threedimensions (Volumetric expansion) while liquids and gases usually suffers change in volume only.

l Linear expansion :

l = l0 (1 + aDq) Þ Dl = l0aDq

0 0+ =D

Before heating After heatingT T+ TD

l Superficial (areal) expansion :

l0 l

ll0

T+ TD

A0 A

TA = A0 (1 + bDq)

Also A0 = l02 and A = l2

So l2 = l0

2(1 + bDq) = [l0(1 + aDq)]2 Þ b =2a

l Volumetric expansion :

l0

ll

l

l0

T+ TDT

l0V0 V

V = V0 (1 + gDq) Also V = l3 and V0=l03 so g =3a

l3=[l0(1+aDq)]3 Þ 6a = 3b = 2g or a : b : g = 1 : 2 : 3

a = coefficient of linear expansion

b = coefficient of superficial expansion

g = coefficient of volumetric expansion

l Contraction on heating :

Some rubber like substances contract on heating because transverse vibration of atoms of substance dominateover longitudinal vibration which is responsible for expansion.

Applications of thermal Expansion in Solids

(a) Bi–metallic strip : When two strips of equal length but of different materials (different coefficientof linear expansion) are joined together, it is called "Bi–metallic strip" and can be used in thermostatto break or make electrical contact. This strip has the characteristic property of bending on heatingdue to unequal linear expansion of the two metals. The strip will bend with metal of greater a on outerside. Coefficient of expansion is more for brass than steel.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

4 E

Pre-Medical : Physics ALLEN

ON OFFBimetallic

strip

At room temperature At higher temperature

SteelSteel

Brass

Room temperature Higher temperature

Brass

(b) Effect of temperature on the time period of a simple pendulum : A pendulum clock keepsproper time at temperature q. If temperature is increased to q' ( > q) then due to linear expansion,length of pendulum increases and hence its time period will increase.

Fractional change in time period DTT

=12

aDq T 1

( T )T 2

D Dµ \ =

l

Q l

l

• Due to increment in its time period, a pendulum clock becomes slow in summer and will lose

time. Loss of time in a time period DT=12

a DqT

• The clock will lose time i.e. will become slow if q' > q (in summer) and will gain time i.e willbecome fast if q' < q (in winter).

• Since coefficient of linear expansion (a) is very small for invar, hence pendulums are made ofinvar to show the correct time in all seasons.

(c) When a rod whose ends are rigidly fixed so as to prevent expansion or contraction undergoes a changein temperature, due to thermal expansion or contraction, a compressive or tensile stress is developedin it. Due to this thermal stress the rod will exert a large force on the supports. If the change in temperatureof a rod of length L is Dq then :–

Heating Cooling

Thermal strain =DLL

= aDq L 1

LD

a = ´Dq

Q So thermal stress = YaDq Q Ystressstrain

=

So force on the supports F=YAaDq

(d) Expansion of cavity: Thermal expansion of an isotropic object may be imagined as a photographicenlargement.

b b

a a

Expansion of A = Expansion of BExpansion of C = Expansion of D

C

rA DBr

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

5E

Pre-Medical : PhysicsALLEN(e) Some other applications

• When rails are laid down on the ground, space is left between the ends of two rails

• The transmission cable are not tightly fixed to the poles

• Test tubes, beakers and cubicles are made up of pyrex–glass or silica because they have verylow value of coefficient of linear expansion

• The iron rim to be put on a cart wheel is always of slightly smaller diameter than that of wheel

• A glass stopper jammed in the neck of a glass bottle can be taken out by warming the neckof the bottle.

l Thermal Expansion in Liquids

• Liquids do not have linear and superficial expansion but these only have volumetric expansion.

• Since liquids are always to be heated along with a vessel which contains themso initially on heating the system (liquid + vessel), the level of liquid in vesselfalls (as vessel expands more since it absorbs heat and liquid expands less)but later on, it starts rising due to faster expansion of the liquid.

QP ® represents expansion of vessel

RPQ

QR ® represents the real expansion of liquid.

• The actual increase in the volume of the liquid

= The apparent increase in the volume of liquid + the increase in the volume of the vessel.

• Liquids have two coefficients of volume expansion.

(i) Co–efficient of apparent expansion (ga)

It is due to apparent (that appears to be, but not in real) increase in the volume of liquid if expansion ofvessel containing the liquid is not taken into account.

Dg = =

´ Dq ´ Dqa

Apparent expansion in volume ( V)Initial volume V

(ii) Co–efficient of real expansion (gr)

It is due to the actual increase in volume of liquid due to heating.

r

Real increase in volume ( V)Initial volume V

Dg = =

´ Dq ´ Dq

• Also coefficient of expansion of flask VesselVessel

( V)VD

g =´ Dq

gReal = gApparent + gVessel

• Change (apparent change) in volume in liquid relative to vessel is

DVapp= V(gReal – gVessel) Dq = V(gr – 3a)Dq

a = Coefficient of linear expansion of the vessel.

• Different level of liquid in vessel

g Dg > g = a Þ g > Dg < g = a Þ g < Dg = g = a Þ g = D =

Real. Vessel app app

Real. Vessel app app

Real Vessel app app

V Level

( 3 ) 0 V ispositive Levelof liquid in vessel will riseonheating

( 3 ) 0 V isnegative Levelof liquid in vessel will fall onheating

( 3 ) 0 V 0 Levelof liquid in vessel will remainsame

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

6 E


Absolute zero

K C F

20.7 -252.5 - 422.5

195 - 78 - 109

273.15 0 32

0 -273.15 - 489.67

273.16 0.01300 27 80.6310.2 37.0 98.6

373.15 100 212Water boilsbody temp.

Room temp.

Hydrogen boils

Solid CO2

Triple point of waterWater freezes

• Actually thermal expansion is always 3-D expansion. When other two dimensions of object are negligiblewith respect to one, then observations are significant only in one dimension and it is known as linear expansion.

• Relation between a , b and g

(i) For isotropic solids: a : b : g = 1 : 2 : 3 or 1 2 3a b g

= =

(ii) For non-isotropic solids b = a1 + a2 and g = a1 + a2 + a3 . Here a1 , a2 and a3 are coefficientof linear expansion in X, Y and Z direction.


A rectangular plate has a circular cavity as shown in the figure. If we increase its temperature then which dimension will increase in following figure.

Solution :Distance between any two point on an object increases with increase in temperature.So, all dimension a, b, c and d will increase

Illustration 7.A small ring having small gap is shown in figure on heating what will happen

to the size of gap.

Solution :

Gap will also increase. The reason is same as in above example.

Illustration 8.

What is the percentage change in length of 1m iron rod if its temperature changes by 100°C.

a for iron is 2 × 10–5/°C.

Solution :

percentage change in length due to temperature change

%l = Dl

l× 100 = aDq × 100 = 2 × 10–5 × 100 × 100 = 0.2%

Illustration 9.

A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length ofthe slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concreteis 1.0 × 10–5 /°C.

Solution :

lt = 10(1 + 1 × 10–5 × 35) = 10.0035 m

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

7E

Pre-Medical : PhysicsALLENIllustration 10.

A glass vessel of volume 100 cm3 is filled with mercury and is heated from 25°C to 75°C. What volumeof mercury will overflow? Coefficient of linear expansion of glass = 1.8 × 10–6/°C and coefficient of volumeexpansion of mercury is 1.8 × 10–4/°C.

Solution :

DV = V0(gL – gC) DT = 100 × {1.8 × 10–4 – 3 × 1.8 × 10–6 } × 50 Þ DV = 0.87 cm3 Ans.

Illustration 11.

There are two spheres of same radius and material at same temperature but one being solid while the otherhollow. Which sphere will expand more if they are heated to the same temperature,

Solution

As thermal expansion of isotropic solids is similar to true photographic enlargement, VV

expansion of a cavity is same as if it had been a solid body of the same material

i.e. DV = Vg Dq

As here V, g and Dq are same for both solid and hollow spheres treated (cavity) ; so the expansion of bothwill be equal.

Illustration 12.

A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is1.2 × 10–5/°C and its Young’s modulus is 2.0 × 1011 N/m2.

Solution : Here final length is less than the original length so that strain is tensile and tensile force is given by

F = AYaDq = 0.5 × 10–6 × 2 × 1011 × 1.2 × 10–5 × 20 = 24 N

BEGINNER'S BOX-1

1. Write down the following temperatures in the increasing order 50°F, 50°C and 50 K.

2. The figure shows three temperature scales with the freezing and boiling points of water indicated.

(a) Rank the size of a degree on these scales, greatest first.

(b) Rank the following temperatures, highest first : 50°X, 50°W and 50°Y.

3. What is the temperature at which we get the same reading on both the Centigrade and Fahrenheit scales?

4. A thin copper wire of length L increases in length by 1% when heated from temperature T1 to T2. What isthe percentage change in area when a thin copper plate having dimensions 2L × L is heated from T1 to T2?

(A) 1% (B) 2% (C) 3% (D) 4%

5. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the changein the diameter of the hole when the sheet is heated to 227 °C ?

Coefficient of linear expansion of copper = 1.70 × 10–5 °C–1,

6. If a bimetallic strip is heated, it will

(A) bend towards the metal with lower thermal expansion coefficient.

(B) bend towards the metal with higher thermal expansion coefficient.

(C) twist itself into helix.

(D) have no bending.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

8 E

Pre-Medical : Physics ALLENHEAT

When a hot body is placed in contact with a cold one, the former gets colder and the latter warmer. From

this observation it is natural to conclude that a certain quantity of heat has passed from the hot body to the

cold one. Heat is a form of energy.

Heat is felt by its effects. Some of the effects of heat are :

(a) Change in the degree of hotness (b) Change in length, surface area and volume

(c) Change in state of a substance (d) Change in the resistance of a conductor

(e) Thermo e.m.f. effect

SI Unit : J (joule). Also measured in calorie.

l Calorie

It is defined as the amount of heat required to raise the temperature of 1 g water by 1°C or 1 K.

l International calorie

International calorie is the amount of heat required to raise the temperature of 1g water from

14.5 °C to 15.5 °C rise of temperature at pressure of 1 atm.

l kilo Calorie

kilo calorie is defined as the amount of heat required to raise the temperature of 1 kg water from 14.5 °C

to 15.5 °C. (1 kcal = 1000 calorie) at pressure of 1 atm.

l British thermal unit (B.T.U.)

It is the amount of heat required to raise the temperature of one pound water by 1°F. (1 B.T.U. = 252 calorie).

Mechanical equivalent of heat

According to Joule, work may be converted into heat and vice–versa. The ratio of work done to the heat

produced is always constant. WH

= constant (J) Þ W = J H

W must be in joule, irrespective of nature of energy or work and H must be in calorie.

J is called mechanical equivalent of heat. It is not a physical quantity but simply a conversion factor.

It converts unit of work into that of heat and vice–versa.

J = 4.186 joule/cal or 4.186 × 10³ joule/kcal. For rough calculations we take J = 4.2 joule/cal

Specific heat (s or c )

The amount of energy required to raise the temperature of unit mass of a substance by 1°C (or 1K) is called

its specific heat. It is represented by s or c.

If the temperature of a substance of mass m changes from T to T + dT when it exchanges an amount of

heat dQ with its surroundings then its specific heat is 1 dQ

c =m dT

The specific heat depends on the pressure, volume and temperature of the substance.

For liquids and solids, specific heat measurements are most often made at a constant pressure as functions

of temperature, because constant pressure is quite easy to produce experimentally.

Unit : joule/kg–K, cal/g –°C

Specific heat of water : cwater = 1 cal/g–°C = 1 cal/g–K = 1 kcal/kg–K = 4200 joule/kg–K

When a substance does not undergo a change of state (i.e., liquid remains liquid or solid remains solid), then

the amount of heat required to raise the temperature of mass m of the substance by an amount Dq is

Q = msDq

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

9E

Pre-Medical : PhysicsALLENVariation in specific heat of water is less than 1% over the interval from 0 to 100°C. Such a small variation

is typical for most solids and liquids, so their specific heats can generally be taken to be constant over fairly

large temperature ranges.

l There are many possible processes to give heat to a gas.

A specific heat can be associated to each such process which depends on the nature of process.

l Value of specific heats of gas can vary from zero (0) to infinity.

l Generally two types of specific heats are defined for a gas –

(a) Specific heat at constant volume (Cv) (b) Specific heat at constant pressure (CP)

l These specific heats can be molar or gram.

Molar heat capacity

The amount of energy needed to raise the temperature of one mole of a substance by 1°C (or 1K) is called

molar heat capacity. The molar heat capacity is the product of molecular weight and specific heat i.e.,

Molar heat capacity C = Molecular weight (M) × Specific heat( c) Þ 1 dQ

Cµ dT

æ ö= ç ÷è ø

If the molecular mass of the substance is M and the mass of the substance is m then number of moles of

the substance µ = mM

Þ C = M dQm dT

æ öç ÷è ø

SI Unit : J/mol–K

Thermal capacity (Heat capacity)

The quantity of heat required to raise the temperature of the whole substance through 1°C is called thermal

capacity. The thermal capacity of mass m of the whole substance of specific heat (s) = ms

Thermal capacity = mass × specific heat

Thermal capacity depends on property of material of the body and mass of the body.

SI Unit : cal/°C or cal/K, Dimensions : ML2 T–2K–1

Water equivalent of a body

As the specific heat of water is unity so the thermal capacity of a body (ms) represents its water equivalent also.

l Mass of water having the same thermal capacity as the body is called the water equivalent of the body

l The water equivalent of a body is the amount of water that absorbs or gives out the same amount of heat

as is done by the body when heated or cooled through 1°C.

Water equivalent= mass of body × specific heat of the material Þ (w = ms).

Latent heat or Hidden heat

When phase of a body changes, change of phase takes place at constant temperature [melting point or boiling

point] and heat released or absorbed is Q = mL where L is latent heat. Heat is absorbed if solid converts

into liquid (at melting point) or liquid convert into vapours (at boiling point) and heat is released if liquid converts

into solid or vapours convert into liquid.

l Latent heat of fusion

It is the quantity of heat (in kilocalories) required to change 1 kg mass of a substance from solid to liquid

state at its melting point. Latent heat of fusion for ice : 80 kcal/kg = 80 cal /g.

l Latent heat of vaporization

The quantity of heat required to change its 1 kg mass from liquid to vapour state at its boiling point.

Latent heat of vaporisation for water : 536 kcal/kg = 536 cal/g ; 540 cal/g

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

10 E

Pre-Medical : Physics ALLENChange of State

l Melting

Conversion of solid into liquid state at constant temperature is known as melting.

l BoilingEvaporation within the whole mass of the liquid is called boiling. Boiling takes place at a constant temperatureknown as boiling point. A liquid boils when the saturated vapour pressure on its surface is equal to atmosphericpressure. Boiling point reduces on decreasing pressure.

l EvaporationConversion of liquid into vapours at all temperatures is called evaporation. It is a surface phenomenon. Greaterthe temperature, faster is the evaporation. Smaller the boiling point of liquid, more rapid is the evaporation.Smaller the humidity, more is the evaporation. Evaporation increases on decreasing pressure that is whyevaporation is faster in vacuum.

l Heat of evaporationHeat required to change unit mass of a liquid into vapour at a given temperature is called heat of evaporationat that temperature.

l SublimationDirect conversion of solid into vapour state is called sublimation.

l Heat of sublimationHeat required to change unit mass of solid directly into vapours at a given temperature is called heat of sublimationat that temperature.

l Camphor and ammonium chloride sublimates on heating in normal conditions.

l A block of ice sublimates into vapours on the surface of moon because of very–very low pressure on its surface

l CondensationThe process of conversion from gaseous or vapour state to liquid state is known as condensation .

These materials again get converted to vapour or gaseous state on heating.

l Hoar frost

Direct conversion of vapours into solid is called hoar frost. This process is just reverse of the process of sublimation.

Ex. : Formation of snow by freezing of clouds.

l Regelation

Regelation is the melting of ice caused by pressure and its resolidification when the pressure is removed.Ice shrinks when it melts, and if pressure is applied, deliberately promoting shrinkage, it is found that meltingis thereby assisted. In other words, melting of cold ice is ordinarily effected by raising the temperature, butif pressure is present to help with the shrinkage the temperature need not be raised so much.

Ice heals up after being cut through by the wire. Melting takes place under the wire because pressurelowers the melting temperature. Refreezing (regelation) occurs above the wire when the water escapes to normalpressure again.

Increase of pressure lowers the melting (or freezing) point of water. Conversely, if a substance expands onmelting, the melting point is raised by pressure.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

11E

Pre-Medical : PhysicsALLENPhase of a substance

The phase of a substance is defined as its form which is homogeneous, physically distinct and mechanicallyseparable from the other forms of that substance.

Phase diagram

l A phase diagram is a graph in which pressure (P) is represented along the y–axis and temperature (T) is representedalong the x–axis.

l Characteristics of Phase diagram

(i) Different phases of a substance can be shown on a phase diagram.

(ii) A region on the phase diagram represents a single phase of the substance, a curve represents equilibriumbetween two phases and a common point represents equilibrium between three phases.

(iii) A phase diagram helps to determine the condition under which the different phases are in equilibrium.

(iv) A phase diagram is useful for finding a convenient way in which a desired change of phase can beproduced.

Phase diagram for water

The phase diagram for water consists of three curves AB, AC and AD meeting each other at the point A,these curves divide the phase diagram into three regions.

Liquid

VapourTriple point

Normal boiling pointNormal melting point

101

0.61

Solid

0.00 0.01 100.00t (°C)c

A

D

B C

Phase diagram for water (Scale are not linear)

At the triple point solid,liquid and vapour phases

all are in equilibrium

fusion curve or ice line

vaporisation curve or steam line

sublimation curve or hoar frost line

P(k

Pa)

(MP) (BP)

P

P

Region to the left of the curve AB and above the curve AD represents the solid phase of water ( ice). Theregion to the right of the curve AB and above the curve AC represents the liquid phase of water. The regionbelow the curves AC and AD represents the gaseous phase of water (i.e. water vapour). A curve on the phasediagram represents the boundary between two phases of the substance.

Along any curve the two phases can coexist in equilibrium

l Along curve AB, ice and water can remain in equilibrium.This curve is called fusion curve or ice line. Thiscurve shows that the melting point of ice decreases with increase in pressure.

l Along the curve AC, water and water vapour can remain in equilibrium. The curve is called vaporisation curveor steam line. The curve shows that the boiling point of water increases with increase in pressure.

l Along the curve AD, ice and water vapour can remain in equilibrium.

This curve is called sublimation curve or hoar frost line.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

12 E

Pre-Medical : Physics ALLENTriple point of water

The three curves in the phase diagram of water meet at a single point A, which is called the triple point

of water. The triple point of water represents the co–existance of all the three phases of water ice water and

water vapour in equilibrium. The pressure corresponding to triple point of water is 6.03 × 10–3 atmosphere

or 4.58 mm of Hg and temperature corresponding to it is 273.16K.

l Significance of triple point of water

Triple point of water represents a unique condition and it is used to define the absolute temperature. While

making Kelvin's absolute scale, upper fixed point is 273.16 K and lower fixed point is 0 K. One kelvin of

temperature is 1

273.16 part of the temperature of triple point of water..

Effect of change in pressure on M.P. and B.P. for water

l If P then result¾¾¾¾¾® M.P. ¯ & B.P.

l If P ¯ then result¬¾¾¾¾¾ M.P. & B.P. ¯

Example :

(1) A bottle is filled with water at 30°C on opening at moon then water will boil and vapourised

(2) At higher altitudes of mountain, food can not be cooked properly.

Heating curve

If to a given mass (m) of a solid, heat is supplied at constant rate and a graph is plotted between temperature

and time, as shown in figure, it is called heating curve.

heat

ing

heating

ofliq

uid

heating of gas

ofso

lid

melting

boiling

A B

C D

a

b

g

tem

prat

ure

time

(heat is suplied at constant rate )

t1 t2 t3 t4

boiling point

melting point

a b g >>(Hea ting ca pac ity)vapour > ( )liquid > ( )solidHe a ting capac ity Hea ting capac ity

slope = dydx

dTdt=

dQ = ms dTmL vapourisation

mL fusion

tan tan tana b g > >

T2

T1

Slope of curve

E

l In the region OA

Rate of heat supply P is constant and temperature of solid is changing with time.

So, Q = mcS DTÞ P Dt = mcS DT [Q Q = P Dt]

Q Tt

DD

= The slope of temperature–time curve so specific heat of solid cS µ 1

slope of line OAspecific heat

(or thermal capacity) is inversely proportional to the slope of temperature–time curve.

l In the region AB

Temperature is constant, so it represents change of state, i.e., melting of solid at melting point T1. At point

A melting starts and at point B all solid is converted into liquid. So between A and B substance is partly solid

and partly liquid. If LF is the latent heat of fusion then

Q = mLF Þ 2 1F

P(t t )L

m-

= [as Q = P(t2 – t1] Þ LF µ length of line AB

i.e., Latent heat of fusion is proportional to the length of line of zero slope.

[In this region specific heat µ 1

tan0°= ¥]

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

13E

Pre-Medical : PhysicsALLENl In the region BC

Temperature of liquid increases so specific heat (or thermal capacity) of liquid will be inversely proportional

to the slope of line BC, L

1c

slope of line BCµ

l In the region CD

Temperature is constant, so it represents change of state, i.e., liquid is boiling at boiling point T2. At C allsubstance is in liquid state while at D is vapour state and between C and D partly liquid and partly gas. Thelength of line CD is proportional to latent heat of vaporisation, i.e., LV µ Length of line CD.

[In this region specific heat µ 1

tan0° = ¥ ]

l In the region DE

The line DE represents gaseous state of substance with its temperature increasing linearly with time. Thereciprocal of slope of line will be proportional to specific heat or thermal capacity of substance in vapourstate.

Law of Mixtures : Principle of calorimetry

l When two bodies at different temperatures are mixed, heat will be transferred from body at higher temperatureto a body at lower temperature till both acquire same temperature. The body at higher temperature releasesheat while body at lower temperature absorbs it, so that

(Principle of calorimetry)

Principle of calorimetry represents the law of conservation of heat energy.

l Temperature of mixture (T) is always ³ lower temperature (TL) and £ higher temperature (TH), L HT T T£ £

The temperature of mixture can never be lesser than lower temperature (as a body cannot be cooled belowthe temperature of cooling body) and greater than higher temperature (as a body cannot be heated abovethe temperature of heating body). Further more usually, rise in temperature of one body may not be equalto the fall in temperature of the other body though heat gained by one body is equal to the heat lost by theother.


5 g ice at 0°C is mixed with 5 g of steam at 100°C . What is the final temperature?

Solution

Heat required by ice to raise its temperature to 100°C,

Q1 = m1L1 + m1c1Dq1 = 5 × 80 + 5 × 1 × 100 = 400 + 500 = 900 cal

Heat given by steam when condensed Q2 = m2L2 = 5 × 536 = 2680 cal

As Q2 > Q1. This means that whole steam is not even condensed.

Hence temperature of mixture will remain at 100°C.

Illustration 14.

A calorimeter of heat capacity 100 J/K is at room temperature of 30°C. 100 g of water at 40°C of specificheat 4200 J/kg–K is poured into the calorimeter. What is the temperature of water in calorimeter?

Solution

Let the temperature of water in calorimeter is t. Then heat lost by water = heat gained by calorimeter

(0.1) × 4200 × (40 – t) = 100 (t – 30) Þ 420 × 40 – 420 t = 100 t – 3000 Þ t = 38.07 °C

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

14 E

Pre-Medical : Physics ALLENIllustration 15.

Find the quantity of heat required to convert 40 g of ice at –20°C into water at 20°C.

Given Lf = 0.336 × 106 J/kg. Specific heat of ice = 2100 J/kg–K, specific heat of water = 4200 J/kg–K

Solution

Heat required to raise the temperature of ice from –20°C to 0°C = 0.04 × 2100 × 20 = 1680 J

Heat required to convert the ice into water at 0°C = mL = 0.04 × 0.336 × 106 = 13440 J

Heat required to heat water from 0°C to 20°C = 0.04 × 4200 × 20 = 3360 J

Total heat required = 1680 + 13440 + 3360 = 18480 J

Illustration 16.

Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at15°C till the temperature of the calorimeter and its contents rises to 80°C. What is the mass of steam condensed?Latent heat of steam = 536 cal/g.

Solution

Heat required by (calorimeter + water)

Q = (m1c1 + m2c2) Dq = (0.02 + 1.1 × 1) (80 – 15) = 72.8 kcal

If m is mass of steam condensed, then heat given by steam

Q = mL + mc Dq = m × 536 + m × 1 × (100 – 80) = 556 m \ 556 m = 72.8

\ Mass of steam condensed m = 72.8

556= 0.130 kg

Illustration 17.

An iron block of mass 2 kg, fall from a height 10 m. After colliding with the ground it loses 25% energyto surroundings. Then find the temperature rise of the block. (Take specific heat of iron 470 J/kg °C)

Solution :

mSDq = 34

mgh Þ Dq = 3 10 10

0.159 C4 470´ ´

= °´

Illustration 18.

The temperature of equal masses of three different liquids A, B, and C are 10°C 15°C and 20°C respectively.The temperature when A and B are mixed is 13°C and when B and C are mixed, it is 16°C. What will bethe temperature when A and C are mixed?

Solution :

when A and B are mixed

mS1 × (13 – 10) = m × S2 × (15 – 13)

3S1 = 2S2 .....(1)

when B and C are mixed

S2 × 1 = S3 × 4 ......(2)

when C and A are mixed

S1(q – 10) = S3 × (20 – q) ......(3)

by using equation (1), (2) and (3)

we get q = 14011

°C

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

15E


5 kg of steam at 100°C is mixed with 10 kg of ice at 0°C. Choose incorrect alternative(Given swater = 1 cal/g°C, LF = 80 cal/g, LV = 540 cal/g)(A) Equilibrium temperature of mixture is 160°C(B) Equilibrium temperature of mixture is 100°C

(C) At equilibrium, mixture contains 1313

kg of water

(D) At equilibrium, mixture contains 123

kg of steam

Solution Ans. (A)Required heat Available heat10 kg ice (0°C) 5 kg steam (100°C)

800 kcal 2700 Kcal

10 kg water (0°C) 5 kg water (100°C)

1000 kcal

10 kg water (100°C)So available heat is more than required heat therefore final temperature will be 100°C.

Mass of heat condensed =+

=800 1000 10

540 3kg. Total mass of water = + = =

10 40 110 13 kg

3 3 3

Total mass of steam = - = =10 5 2

5 1 kg3 3 3

BEGINNER'S BOX-2

1. A bullet of mass 10 g is moving with speed 400m/s. Find its kinetic energy in calories ?

2. Calculate amount of heat required to convert 1 kg steam from 100°C to 200°C steam ?

3. Calculate heat required to raise the temperature of 1 g of water by 1°C ?

4. 420 J of energy supplied to 10 g of water will raise its temperature by ?

5. The ratio of the densities of the two bodies is 3 : 4 and the ratio of specific heats is 4 : 3 . Find the ratio

of their thermal capacities for unit volume ?

6. Heat releases by 1 kg steam at 150°C if it is converted into 1 kg water at 50°C.

7. 200 g water is filled in a calorimetry of negligible heat capacity. It is heated till its temperature is increase

by 20°C. Find the heat supplied to the water.

8. A bullet of mass 5 gm is moving with speed 400 m/s strike a target. Then calculate rise of temperature of

bullet. Assuming all the lose in kinetic energy is converted into heat energy of bullet if its specific heat is

500J/kg°C.

9. 1 kg ice at –10°C is mixed with 1 kg water at 100°C. Then find equilibrium temperature and mixture content.

10. 540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is

(Given latent heat of fusion of ice = 80 cal/g and specific heat capacity of water = 1 cal/g0C)

(A) 0°C (B) 40°C (C) 80°C (D) less than 0°C

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

16 E

Pre-Medical : Physics ALLENMODES OF HEAT TRANSFER

Heat is a form of energy which transfers from a body at higher temperature to a body at lower temperature.

The transfer of heat from one body to another may take place by any one of the following modes :.

l Conduction

The process in which the material takes an active part by molecular action and energy is passed from oneparticle to another is called conduction. It is predominant in solids.

l Convection

The transfer of energy by actual motion of particle of medium from one place to another is called convection.It is predominant is fluids (liquids and gases).

l Radiation

Quickest way of transmission of heat is known as radiation. In this mode of energy transmission, heat is transferredfrom one place to another without effecting the inter–venning medium.

Conduction Convection Radiation

Heat Transfer due to Temperature difference

Heat transfer due to density difference

Heat transfer without any medium

Due to free electron or vibration motion of molecules

Actual motion of particles Electromagnetic radiation

Heat transfer in solids and Hg

Heat transfer in fluids (Liquid + gas)

No medium required

Slow process Slow process Fast process (3 × 108 m/sec)

Irregular path Irregular path Straight line (like light)

3.1 Thermal conduction

The process by which heat is transferred from hot part to cold part of a body through the transfer of energy

from one particle to another particle of the body without the actual movement of the particles from their

equilibrium positions is called conduction. The process of conduction takes place only in solid body (except

Hg). Heat transfer by conduction from one part of body to another continues till their temperatures become

equal.

For example if you hold an iron rod with one of its end on a fire for some time, the handle will become

hot. The heat is transferred from the fire to the handle by conduction along the length of iron rod. The vibrational

amplitude of atoms and electrons of the iron rod at the hot end takes relatively higher values due to the higher

temperature of their environment. These increased vibrational amplitude are transferred along the rod, from

atom to atom during collision between adjacent atoms. In this way a region of rising temperature extends

itself along the rod to your hand.

TC

Q2

Q1

TH

Ox dx

BA

L

Consider a slab of face area A, Lateral thickness L, whose faces have temperatures TH and TC(TH > TC).

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

17E

Pre-Medical : PhysicsALLENLet temperature of face A be T and that of face B be T + DT. Then experiments show that Q, the amountof heat crossing the area A of the slab at position x in time t is given by

Q dTKA

t dx= - or H CKA(T T )Q

t L-

=

K ® Thermal conductivity

® It is the measure of the ability of material to conduct the heat.

Here K is a constant depending on nature of the material of the slab and is named thermal conductivity of

the material, and the quantity æ öç ÷è ø

dTdx

is called temperature gradiant. The (–) sign shows heat flows from high

temperature to low temperature (DT is a –ve quantity).

Steady State

If the temperature of a cross-section at any position x in the above slab remains constant with time (remember,it does vary with position x), the slab is said to be in steady state and temperature of rod is not same.

Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in theslab must be same.

For a conductor in steady state there is no absorption or emission of heat at any cross-section. (as temperatureat each point remains constant with time). The left and right faces are maintained at constant temperaturesTH and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes throughthem and same amount of heat flows through each cross-section in a given interval of time.

Hence Q1 = Q = Q2. Consequently the temperature gradient is constant throughout the slab.

Hence,--D

= = = C Hf i T TT TdT Tdx L L L

andD

= -Q T

KAt L

Þ tQ

= KA-æ ö

ç ÷è øH CT T

L

Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval t.

Thermal conductivity (K) :

• It depends on nature of material.

Order of thermal conductivity Ag > Cu > Au > Al For Ag maximum is (410 W/mK)

For Freon minimum is 12 (0.008 W/mK)K

• SI Unit : J s–1 m–1 K–1 Dimensions : M1 L1 T–3 q–1

• For an ideal or perfect conductor of heat the value of K = ¥

• For an ideal or perfect bad conductor or insulator the value of K = 0

• For cooking the food, low specific heat and high conductivity utensils are most suitable.

Application of Thermal Conduction

• In winter, the iron chairs appear to be colder than the wooden chairs.

• Cooking utensils are made of aluminium and brass whereas their handles are made of wood.

• Ice is covered in gunny bags to prevent melting of ice.

• We feel warm in woollen clothes and fur coat.

• Two thin blankets are warmer than a single blanket of double the thickness.

• Birds often swell their feathers in winter.

• A new quilt is warmer than old one.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

18 E

Pre-Medical : Physics ALLENThermal Resistance to conduction

If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in

your tiffin-box, you are more interested in poor heat conductors, rather than good conductors. For this reason,

the concept of thermal resistance R has been introduced.

For a slab of cross-section A, Lateral thickness L and thermal conductivity K, =L

RKA

In terms of R, the amount of heat flowing though a slab in steady-state (in time t)-

= H L(T T )Qt R

If we name Qt

as thermal current iT then, H LT

T Ti

R-=

This is mathematically equivalent to OHM’s law, with temperature doing the role of electric potential. Hence

results derived from OHM’s law are also valid for thermal conduction.

More over, for a slab in steady state we have seen earlier that the thermal current iL remains same at each

cross-section. This is analogous to Kirchhoff’s current law in electricity, which can be very conveniently applied

to thermal conduction.

l Equivalent conductivity for Heat flow through slabs in series

Req = R1 + R2

L LK Aeq

1 2+ =

LK A

1

1

+L

K A2

2

K1 K2

L1 L2

T1 T0 T2

QtQ

t

R 1 R2

equivalent to

Equivalent thermal conductivity of the system is

Keq = L LLK

LK

1 2

1

1

2

2

+

+ =

S

S

LLK

i

i

i

l Equivalent thermal condctivity for Heat flow through slabs in parallel

R = L

KA

1 1 1

1 2R R Req

= +

K

Leq

(A1 + A2) = K A

LK A

L1 1 2 2+

Equivalent thermal conductivity KK A K A

A Aeq =++

1 1 2 2

1 2

= S

SK AAi i

i

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

19E

Pre-Medical : PhysicsALLENGrowth of Ice on LakesIn winter atmospheric temperature falls below 0°C and water in the lake start freezing. Let at time t thicknessof ice on the surface of the lake = x and air temperature = –q° C

The temperature of water in contact with the lower surface of ice = 0°C

Let area of the lake = A

Heat escaping through ice in time dt is dQ KAx

dt=- -[ ( )]0 q

water

ice xdx

air at below 0° C

0° CDue to escape of this heat increasing extra thickness of ice = dx

Mass of this extra thickness of ice is m = rV = r A.dx

dQ = mL = (r A.dx) L

\ KAx

dtq

= (r A.dx) L Þ dtL

Kx dx=

rq

So time taken by ice to grow a thickness x is t = r

qr

qL

Kx dx

LK

xx

0

12

2z =

So time taken by ice to grow from thickness x1 to thickness x2 is

t = t2 – t1 = 12

rLKT

(x22 – x1

2) and 2 22 1t (x x )µ -

Time taken to double and triple the thickness ratio t1 : t2 : t3 :: 12 : 22 : 32 So t1 : t2 : t3 :: 1 : 4 : 9

3.2 Convection

A

Flame

Convection requires a medium and is the process in which heat is transferred fromone place to other by actual movement of heated substance (usually fluid).The typeof convection which results from difference in densities is called natural convection (forexample, a fluid in a container heated through its bottom). However, if a heated fluidis forced to move by a blower, fan or pump, the process is called forced convection.

Phenomena Based on convection :(i) Land and sea breezes :

The heat from the Sun is absorbed more rapidly by land than by sea–water. Moreover,the specific heat of land is low as compared to that of sea–water. Consequently, therise in temperature of land is higher as compared to that of sea–water. To sum–up,land is hotter than the sea during day time. As a result of this, the colder air over thesea blows towards the land. This is called sea–breeze.

At night, air blows from land towards sea. This is called land breeze.

(ii) Formation of trade winds :The surface of Earth near the equator gets heated strongly. So, the air in contact with the surface of Earthat the expands and rises upwards. As a result of this, a low pressure is created at the equator.

At the poles, the air in the upper atmosphere gets cooled and comes down. So, a high pressure is createdat the poles. Due to difference of pressures at the poles and equator, the air at the poles moves towardsthe equator, rises up, moves towards poles and so on. In this way, a wind is formed in the atmosphere.

The rotation of the Earth also affects the motion of the wind. Due to anti–clockwise rotation of Earth thewarm wind blowing from equator to north drifts towards east. The steady wind blowing from north–East toequator, near the surface of Earth, is called trade wind.

(iii) Monsoons :In summer, the peninsular mass of central Asia becomes more strongly heated than the water of the IndianOcean. This is due to the fact that the specific heat of water is much higher than that of the soil and rocks.Hot air from the heated land mass rises up and moves towards the Indian ocean. Air filled with moistureflows over the Indian ocean on the south towards heated land mass. When obstructed by mountains, the moistair rushes upwards to great height. In the process, it gets cooled. Consequently, the moisture condenses andfalls as rain.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

20 E

Pre-Medical : Physics ALLEN(iv) Ventillation :

Ventillator of exhaust fan in a room help of remove impure and warm air from a room. The fresh air fromoutside blows into the room. This is all due to the forced convection current set up in the room.

(v) To regulate temperature in the human body :Heat transfer in the human body involves a combination of mechanisms. These together maintain a remarkablyuniform temperature in the human body inspite of large changes in environmental conditions. The chief internalmechanism is forced convection. The heart serves as the pump and the blood as the circulating fluid.

Some important points :l Natural convection takes place from bottom to top due to gravity while forced convection in any direction.

l In case of natural convection, convection currents move warm air upwards and cool air downwards.This is why heating is done from base, while cooling from the top.

l Natural convection is not possible in a gravity free region such as a freely falling lift or an orbiting satellite.

l Natural convection plays an important role in ventilation, in changing climate and weather and in formingland and sea breezes and trade winds.

l The forced convection of blood in our body by a pump (heart) helps in keeping the temperature ofbody constant.

l Convection takes place in fluids (gases and liquid)

GOLDEN KEY POINTS

l For heat propagation via natural convection, temperature gradient exists in vertical direction and not inhorizontal direction.

l Most of heat transfer that is taking place on Earth is by convection, the contribution due to conduction andradiation is very small.


One face of an aluminium cube of edge 2 metre is maintained at 100 °C and the other end is maintainedat 0 °C. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cubein 5 seconds. (thermal conductivity of aluminium is 209 W/m–°C)

SolutionHeat will flow from the end at 100°C to the end at 0°C.

Area of cross-section perpendicular to direction of heat flow, A = 4m2 then-

= H C(T T )QKA

t L2(209W / m C)(4m )(100 C 0 C)(5sec)

Q 209 kJ2m

° ° - °= =

Illustration 21. 50ºC

4ºC

steel

copper

Aluminium12ºC

Three identical rods of length 1m each, having cross-section area

of 1cm2 each and made of Aluminium, copper and steel respectively

are maintained at temperatures of 12°C, 4°C and 50°C respectively at

their separate ends. Find the temperature of their common junction.

[ KCu=400 W/m-K , KAl = 200 W/m-K , Ksteel = 50 W/m-K ]

Solution

RAl = 4

4

L 1 10KA 200200 10-

= =´

Similarly Rsteel = 410

50 and Rcopper =

410400

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

21E

Pre-Medical : PhysicsALLENLet temperature of common junction = T 50ºC

4ºC

12ºCiAl

RCu

iS

iCu

Rs

RAl T

then from Kirchhoff's current law, iAl + isteel + iCu = 0

ÞAl steel Cu

T 12 T 50 T 40

R R R- - -

+ + =

Þ (T – 12) 200 + (T – 50) 50 + (T – 4) 400 = 0

Þ 4(T – 12) + (T – 50) + 8 (T – 4) = 0

Þ 13T = 48 + 50 + 32 = 130 Þ T = 10°C

Illustration 22.

The thermal conductivity of brick is 1.7 W m–1 K–1, and that of cement is 2.9 W m–1 K–1. What thickness

of cement will have same insulation as the brick of thickness 20 cm ? Assuming their area to be same.

Solution

Since Q = KA T T t

L( )1 2-

. For same insulation by the brick and cement ; Q, A (T1 – T2) and t do not change.

Hence, KL

remain constant. If K1 and K2 be the thermal conductivities of brick and cement respectively and

L1 and L2 be the required thickness then KL

1

1=

2

2

KL or

1720.

=29

2

.L \ L2 =

2917..

× 20 = 34.12 cm

Illustration 23.

Two vessels of different materials are identical in size and wall–thickness. They are filled with equal quantities

of ice at 0°C. If the ice melts completely, in 10 and 25 minutes respectively then compare the coefficients

of thermal conductivity of the materials of the vessels.

Solution

Let K1 and K2 be the coefficients of thermal conductivity of the materials, and t1 and t2 be the time in which

ice melts in the two vessels. Since both the vessels are identical, so A and L in both the cases is same.

Now, Q = K A t

L1 1 2 1( )q q-

= K A t

L2 1 2 2q q-b g

ÞKK

1

2=

tt2

1=

2510

minmin =

52

Illustration 24.

Two plates of equal areas are placed in contact with each other. Their thickness are 2.0 cm and 5.0 cm

respectively. The temperature of the external surface of the first plate is –20°C and that of the external surface

of the second plate is 20°C. What will be the temperature of the contact surface if the plate (i) are of the

same material, (ii) have thermal conductivities in the ratio 2 : 5.

Solution

Rate of flow of heat in the plates is Qt

= K A

L1 1

1

( )q q- =

K AL

2 2

2

( )q q-...(i)

plat

e 1

plate 2

5cm2cm

-20 C0 20 C0q(i) Here q1 = –20°C, q2 = 20°C,

L1 = 2 cm = 0.02 m, L2 = 5 cm = 0.05 m and K1 = K2 = K

\ equation (i) becomesKA - -20

002

qb g.

= KA q - 20

005

b g.

\ 5(–20–q) = 2(q – 20) Þ –100 – 5q = 2q – 40 Þ 7q = –60 Þ q = –8.6°C

(ii)KK

1

2 =

25

or K1 = 25

K2

\ from equation (i) 2 5 20

0022K A - - qb g

.=

K A2 20

005

q -b g.

Þ –20 – q = q – 20 \ q = 0°C

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

22 E


Two thin concentric shells made of copper with radius r1 and r2 (r2 > r1) have a material of thermal conductivityK filled between them. The inner and outer spheres are maintained at temperatures TH and TC respectivelyby keeping a heater of power P at the centre of the two spheres. Find the value of P.

Solution : Heat flowing per second through each cross-section of the sphere = P = i.

Thermal resistance of the spherical shell of radius x and thickness dx,

Pdx

r1

x

r2

dR = 2x4.Kdxp

Þ R = ò p

2

1

r

r2 K.x4

dx = K4

1p

÷÷ø

öççè

æ-

21 r1

r1

thermal current

i = P = RTT CH -

= )rr(rr)TT(K4

12

21CH

--p

.

Illustration 26.

Water in a closed tube is heated with one arm vertically placed above the lamp. In what direction water willbegin the circulate along the tube ?

Solution

On heating the liquid at A will become lighter and will rise up. This will push the A B

liquid in the tube upwards and so the liquid in the tube will move clockwise i.e. fromB to A.

BEGINNER'S BOX-3

1. Explain why :

(a) a brass tumbler feels much colder than a wooden tray on a chilly day

(b) two layers of cloth of equal thickness provide warmer covering than a single layer of clothof double thickness ?

(c) mud-houses are colder in summer and warmer in winter?

(d) In winter birds sit with their wings spread out ?

(e) Woollen clothes are warmer than cotton clothes ?

2. Two metal cubes with 3 cm-edges of copper and aluminium are arranged as shown in figure. Find

(a) The total thermal current from one reservoir to the other.

(b) The ratio of the thermal current carried by the copper cube to that carried by the aluminium cube.Thermal conductivity of copper is 60 W/m–K and that of aluminium is 40 W/m–K.

3. For shown situation, calculate the temperature of the common interface.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

23E

Pre-Medical : PhysicsALLEN4. Calculate q1 and q2 in shown situation.

K 2K 1.5K

200°C 18°Cq1 q2

l l l

5. The temperature at the ends of a uniform rod of length 100 cm are respectively 95°C and 5°C. What willbe the temperature at a point 30 cm far from the hotter end ? Also calculate the temperature gradient.

6. Three conducting rods of same material and cross-section are shown in figure.Temperature of A, D and C are maintained at 20°C, 90°C and 0°C. Find the ratio

of length BD and BC if there is no heat flow in AB.

3.3 Thermal RadiationThe process of the transfer of heat from one place to another place without heating the intervening mediumis called radiation. When a body is heated and placed in vacuum, it loses heat even when there is no mediumsurrounding it. The heat can not go out from the body by the process of conduction or convection sinceboth of these process require the presence of a material medium between source and surrounding objects.The process by which heat is lost in this case is called radiation. This does not require the presence of anymaterial medium.It is radiation by which heat from the Sun reaches the Earth. Radiation has the following properties:(a) Radiant energy travels in straight lines and when some object is placed in the path, its shadow is formed

at the detector.

(b) It is reflected and refracted or can be made to interfere. The reflection or refraction are exactly as incase of light.

(c) It can travel through vacuum.

(d) Intensity of radiation follows the law of inverse square.

(e) Thermal radiation can be polarised in the same way as light by transmission through a nicol prism.

(f) Radiation takes place in solid, liquid & gases.

All these and many other properties establish that heat radiation has nearly all the properties possessed bylight and these are also electromagnetic waves with the only difference of wavelength or frequency. Thewavelength of thermal radiation is larger than that of visible light.

• When radiation passes through any medium then radiations slightly absorbed by medium according to its

absorptive power so temperature of medium slightly increases.

• In order to obtain a spectrum of radiation, a special prism is used like KCl prism, Rock salt prism Flourspar

prism. Normal glass prism or Quartz prism can not be used (because it absorbs some radiation).

• Radiation intensity is measured with a specific device named as Bolometer.

• Heat radiation are always obtained in infra–red region of electromagnetic wave spectrum so they are called

Infra-red rays.

• Thermal radiation when incident on a surface, then exert pressure on the surface which is known as Radiation

Pressure.

Basic Fundamental definitions

• Absorptive power or absorptive coefficient (a) : The ratio of amount of radiation absorbed by a surface

(Qa) to the amount of radiation incident (Q) upon it, is defined as the coefficient of absorption aQa

Q= . It

is unitless and dimensionless.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

24 E


• Spectral absorptive power (al )Qa

a =Q

ll

l : Also called monochromatic absorptive coefficient

At a given wavelength a a d=¥z l l0

. For ideal black body al and a = 1, a and al are unitless

• Emissive power (e) : The amount of heat radiation emitted by unit surface area in unit second at a particular

temperature.SI UNIT : J/m2–s or watt/m2

• Spectral Emmisive power (el) : The amount of heat radiation emitted by unit area of the body in one second

in unit spectral region at a given wavelength.

Emissive power or total emissive power 0

e e d¥

l= lòSI UNIT : W/m2–Å

Emissivity (e)

• Absolute emissivity or emissivity : Radiation energy given out by a unit surface area of a body in unit

time corresponding to unit temperature difference w.r.t. the surroundings is called Emissivity.

S I unit : W/m2 K

• Relative emissivity (er) : er = QQ

GB

IBB

= eE

GB

IBB

= emitted radiation by gray body

emitted radiation by ideal black body

GB = gray or general body, IBB = Ideal black body

(i) No unit (ii) For ideal black body er = 1 (iii) range 0 < er < 1

Spectral, emissive, absorptive and transmittive power of a given body surface

Due to incident radiations on the surface of a body following phenomena occur by which the radiation is

divided into three parts. (a) Reflection (b) Absorption (c) Transmission

• From energy conservation

amount of absorbed

radiation Qa

amount of incident radiation Q

amount of reflectedradiation Qr

amount of transmittedradiation Q t

Q = Qr + Qa + Qt Þ ar tQQ Q1

Q Q Q+ + = Þ r + a + t = 1

Reflective Coefficient r = rQQ

, Absorptive Coefficient a = aQQ

,

Transmittive Coefficient t = tQ

Q

r = 1 and a = 0 , t = 0 Þ Perfect reflector

a = 1 and r = 0, t = 0 Þ Ideal absorber (ideal black body)

t = 1 and a = 0, r = 0 Þ Perfect transmitter (diathermanous)

Reflection power (r) = rQ

100 %Q

é ù´ê úë û, Absorption power (a) =

aQ100 %

Qé ù´ê úë û

Transmission power (t) = tQ

100 %Q

é ù´ê úë û

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

25E

Pre-Medical : PhysicsALLENIdeal Black Body

• A body surface which absorbs all incident thermal radiations at low temperature, irrespective of their wave

length and emits out all these absorbed radiations at high temperature is assumed to be an ideal black body

surface.

• The identical parameters of an ideal black body is given by a = al = 1 and r = 0 = t, er = 1

Ferry's ideal black body

• The nature of emitted radiations from surface of ideal black body only depends on its temperature

• The radiations emitted from surface of ideal black body are called as either full or white radiations.

• At any temperature the spectral energy distribution curve for surface of an ideal black body is always continuous

and according to this concept if the spectrum of a heat source obtained is continuous then it must be ideal

black body like kerosene lamp; oil lamp, heating filament etc.

• There are two experimentally ideal black body

(a) Ferry's ideal black body (b) Wien's ideal black body.

• At low temperature, surface of ideal black body is a perfect absorber and at a high temperature it proves

to be a perfect emitter.

• An ideal black body need not be of black colour (eg. Sun).

Prevost's theory of heat energy exchange

According to Prevost, at every possible temperature ( except zero kelvin temperature) there is a continuous

heat energy exchange between a body and its surrounding and this exchange carry on for infinite time.

The relation between temperature difference of body with its surrounding decides whether the body experience

cooling effect or heating effect.

When a cold body is placed in the hot surrounding : The body radiates less energy and absorbs more

energy from the surrounding, therefore the temperature of body increases. (Heating effect)

When a hot body placed in cooler surrounding : The body radiates more energy and absorbs less energy

from the surroundings. Therefore temperature of body decreases. (cooling effect)

When the temperature of a body is equal to the temperature of the surrounding

The energy radiated per unit time by the body is equal to the energy absorbed per unit time by the body,

therefore its temperature remains constant and the body is in thermal equillibrium with surrounding. Hence

no heating and cooling effects are seen.

KIRCHHOFF'S LAW

At a given temperature for all bodies the ratio of their spectral emissive power (el) to spectral absorptive power

(al) is constant and this constant is equal to spectral emissive power (El) of the ideal black body at same temperature

ea

El

ll= = constant

1 2

e ea a

l l

l l

é ù é ù=ê ú ê ú

ë û ë û= constant Hence e al lµ

Good absorbers are good emitters and bad absorbers are bad emitters

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

26 E

Pre-Medical : Physics ALLENApplications of Kirchoff Law

• Fraunhoffer's lines

Fraunhoffer lines are dark lines in the spectrum of the Sun. When white light emitted from the central core

of the Sun (Photosphere) passes through its atmosphere (chromosphere) some of the radiations are absorbed

by the gases present, resulting in dark lines in the spectrum of Sun. At the time of total solar eclipse direct

light rays emitted from photosphere cannot reach on the Earth and only rays from chromosphere are able

to reach on the Earth surface. At that time we observe bright fraunhoffer lines.

chromosphere

photosphere

SUN

10 K7

6000K

NaCa

vapour state

K

• In deserts days are hot and nights cold

Sand is rough and black, so it is a good absorber and hence in deserts, days (when radiation from Sun is

incident on sand) will be very hot. Now in accordance with Kirchhoff's Law, good absorber is a good emitter.

So nights (when send emits radiation) will be cold.

Stefan's Law

The amount of radiation emitted per second per unit area by ideal black body is directly proportional to the

fourth power of its absolute temperature.

Amount of radiation emitted 4E Tµ where T = temperature of ideal black body (in K)

E = s T4 (This law is true for only ideal black body)

SI Unit : E = watt/m2 s ® Stefen's constant = 5.67 x10–8 watt /m2 K4 (universal constant)

Dimensions of s : M1 L0 T–3 q–4

Total radiation energy emitted out by surface of area A in time t :

Ideal black body QIBB = s A T4 t and for any other body QGB = ersA T4 t

Rate of emission of radiation

When temperature of surrounding T0 (Let T0 < T)

Rate of emission of radiation from per unit area of ideal black body surface E1 = s T4

Rate of emission or absorption of radiation (per unit area) from surrounding E2 = sT04

Net rate of loss of radiation per unit area from ideal black body surface is

E = E1 – E2 = sT4– s T04 = s ( T4 – T0

4 )

Net loss of radiation energy from entire surface area in time t is QIBB = sA ( T4 – T04 ) t

For any other body QGB = er As ( T4 – T04 ) t

If in time dt the net heat energy loss for ideal black body is dQ and because of this its temperature falls by dT

Rate of loss of heat (IBB)4 4

H 0

dQR = A(T T ) J/s

dt= s -

It is also equal to emitted power or radiation emitted per second

Rate of fall in temperature (Rate of cooling) 4 4

F 0

dT AR = (T T )

dt mss

= - dQ dTms

dt dté ù=ê úë ûQ

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

27E

Pre-Medical : PhysicsALLEN

Body RH RF

(i) Two solid sphere (same material) RH µ r2F

1R

rµ

(same T, T0 , s, r)

(different radius r1, r2)

(ii) Two solid sphere

(different material) (same T, T0 ) RH µ r2F

1R

r sµ

r

(iii) Different shape bodies RH µ A F

AR

Vµ

(Cube, sphere, cylinder • maximum for flat surface • maximum for flat surface

flat surface) • minimum for spherical • minimum for spherical surface

(const. T, T0,V, same materials) surface

(iv) Two sphere • RH is same for both RF

(one solid and another hollow) • maximum for hollow sphere

(T, T0, s, A are same) • minimum for solid sphere

(v) Different bodies made • RH is same for all bodiesF

1R

sµ

of different material

(T,T0, M, A)(s-different)

Where T and T0 absolute temperature of body and surrounding, M = mass of body,

s = Specific heat, r = density, V = volume.

l When a body cools by radiation, its cooling depends on :

(i) Nature of radiating surface : greater the emissivity (er), faster will be the cooling.

(ii) Area of radiating surface : greater the area of radiating surface, faster will be the cooling.

(iii) Mass of radiating body : greater the mass of radiating body slower will be the cooling.

(iv) Specific heat of radiating body : greater the specific heat of radiating body slower will be the cooling.

(v) Temperature of radiating body : greater the temperature of radiating body faster will be the cooling.

NEWTON'S LAW OF COOLING

Rate of cooling dTdt

æ öç ÷è ø

is directly proportional to excess of temperature of the body over that of surrounding.

[(when (T – T0) > 35°C]dTdt

- µ (T – T0)

T = temperature of body [all temperatures in °C]

To = temperature of surrounding,

T – T0 = excess of temperature ( T > T0 )

If the temperature of body is decreased by dT in time dt then rate of fall of temperature 0

dT– K(T T )

dt= -

Where negative sign indictates that the rate of cooling is decreasing with time.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

28 E

Pre-Medical : Physics ALLENFor Numerical Problems, Newton's Law of cooling T

T0

t

If the temperature of body decreases from T1 to T2 and temperature

of surroundings is T0 then

average excess of temperature = 1 20

T TT

2+é ù-ê úë û

Þ1 2 1 2

0

T T T TK T

t 2- +é ù é ù= + -ê ú ê úë û ë û

Limitations of Newton's Law of Cooling

• Temperature difference should not exceed 35° C, (T – T0) /> 35° C

• Loss of heat should only be by radiation.

• This law is an extended form of Stefan–Boltzman's law.

Derivation of Newton's law from Stefan's Boltzman law

440

dT A(T T )

dt mss

= - 0

0

T T T

T T T

- = Dì üí ý

= + Dî þ

4 40 0

dT A(T T) T

dt mss é ù= + D -ë û If x <<< 1 then (1 + x)n = 1 + nx (Binomial theorem)

4 4 40 0

0

dT A TT (1 ) T

dt ms Té ùs D

= + -ê úë û

=4 40

0

A TT (1 ) 1

ms Té ùs D

+ -ê úë û

=40

0

A TT 1 4 1

ms Té ùs D

+ -ê úë û

30

dT A4 T T

dt m sé ùs

= Dê úë û

Þ dTK T

dt= D here constant

304 A T

Kms

s=

Newton's law of coolingdT

Tdt

µ D (for small temperature difference)

Spectral Energy distribution curve of Black Body radiations

Practically given by : Lumers and Pringshem Mathematically given by : Plank

T3

T2

T1

lm3lm2

lm1 l

T > T > T3 2 1

E lm1

E lm2

E lm3

lm3lm2

lm1< <

inte

nsity

Jse

c-m

2

different wavelength and emitted radition

Spe

ctra

l rad

iatio

n

T3

T1

T > T > T3 2 1

E n1

E n2

E n3

n3 2 1 > > n n

n1 n2 n3

Enor

I l

T2

n

differentfrequencyand emittedradiation

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

29E

Pre-Medical : PhysicsALLENResults from these Graphs

E l

l

area T4µ

spectral energy distribution curve (E – )l l

height of peak T5µ

dllm

(i) m

1T

l µ Wien's displacement law

mT bl =

1 2m 1 m 2T Tl = l

(ii)m

5E Tl µ

Area between curve and l axis gives (iii) i.e. Area 4

0

E d E T¥

l l = = sòthe emissive power of body

Hence

4

1 1

2 2

A TA T

é ù= ê ú

ë ûWein's Displacement Law

The wavelength corresponding to maximum emission of radiation decrease with increasing

temperaturem

1T

é ùl µê úë û. This is known as Wein's displacement law.

mT bl = where b is Wein's constant = 2.89 x 10–3 mK.

Dimensions of b : = M0 L1 T0 q1

Relation between frequency and temperature m

cT

bn = [ ]c = l ´ n

Solar constant 'S'The Sun emits radiant energy continuously in space of which an in significant part reaches the Earth. Thesolar radiant energy received per unit area per unit time by a black surface held at right angles to the Sun'srays and placed at the mean distance of the Earth (in the absence of atmosphere) is called solar constant.The solar constant S is taken to be 1340 watt/m2 or 1.937 Cal/cm2–minute

l Temperature of the SunLet R be the radius of the Sun and 'd' be the radius of Earth's orbit around

R

A

T

d

A'

the Sun. Let E be the energy emitted by the Sun per second per unit area. Thetotal energy emitted by the Sun in one second = E.A = E × 4pR2. (This energyis falling on a sphere of radius equal to the radius of the Earth's orbit aroundthe Sun i.e., on a sphere of surface area 4pd2)So, The energy falling per unit time per unit area of Earth

= 4

4

2

2

pp

R E

d

´=

E R

d

2

2

R = 7× 108m , d = 1.5 × 1011 m, s = 5.7 × 10–8 W m–2 K–4

Solar constant S = E R

d

2

2

By Stefan's Law E = sT4

S = s T R

d

4 2

2 Þ T =

1 12 11 24 4

2 8 8 2

S d 1340 (1.5 10 )R 5.7 10 (7 10 )-

é ù é ù´ ´ ´=ê ú ê ús ´ ´ ´ ´ë û ë û

= 5732 K

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

30 E


• At absolute zero temperature (zero kelvin) all atoms of a given substance remains in ground state, so, at

this temperature emission and absorption of radiation from any substance is impossible, so Prevost's heat

energy exchange theory does not applied at this temperature, so it is called limiting temperature of prevost's

theory.

• With the help of Prevost's theory rate of cooling of any body w.r.t. its surroundings can be worked out (applied

to Stefan Boltzman law, Newton's law of cooling.)

• For a constant temperature the spectral emmisive power of an ideal black body is a constant parameter

• The practical confirmation of Kirchhoff's law carried out by Rishi apparatus and the main base of this apparatus

is a Lessilie container.

• The main conclusion predicted from Kirchhof's law can be expressed as

Good absorber � Good emitter

Bad absorber � Bad emitter

(at Low temperature) (at high temperature)

• If all of T, T0, m, s, V, r, are same for different shape body then RF and RH will be maximum for the flat

surface.

• If a solid and hollow sphere are taken with all the parameters same then hollow will cool down at fast rate.

• Rate of temperature fall , F

1 dTR

s dtµ µ

so dt µ s. If condition of specific heat is s1 > s2 > s3

and if all cooled same temperature i.e. temperature fall is also identical for all then required time t sµ

\ t1 > t2 > t3• Spectral energy distribution curves are continuous. At any temperature in all possible wavelength

radiation between (0 – ¥) are emitted but quantity of radiations are different for different wavelength.

• As the wave length increases, the amount of radiation emitted first increase, becomes maximum and then

decreases.

• At a particular temperature the area enclosed between the spectral energy curve and wavelength axis shows

the emissive power of the body.

4

0

Area E d E T¥

l= l = = sò


If lm for the moon is 14.5 micron, then find its temperature.

Solution

Wien's displacement law lmT = b

\ T = b

ml =

289 10

145 10

3

6

.

.

´´

-

- = 199.3K

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

31E


Total radiation incident on body is 400 J, If 20% of incident radiation reflected back and 120 J is absorbed

by body. Then find out transmittive power in percentage.

Solution

Q = Qt + Qr + Qa Þ 400 = 80 + 120 + Qt Þ Qt = 200

So transmittive power is tQ100%

Q´ = 50%

Illustration 29.

The operating temperature of a tungesten filament in an incandescent lamp is 2000 K and its emissivity is 0.3.

Find the surface area of the filament of a 25 watt lamp. Stefan's constant s = 5.67 × 10–8 Wm–2 K

–4

Solution

Q Rate of emission = wattage of the lamp

\ W = AesT4 Þ A = W

e Ts 4 = 8 4

250.3 5.67 10 (2000)-´ ´ ´ = 0.918 × 10–4 m2

Illustration 30.

Draw a graph between logE and log T

log T x

log E

log s

y

Solution

E = s T4 (taking log)

4logE = log ( T )s

logE = 4logT +logs This is equivalent to y = mx – C (s < 1 so its log is a negative quantity)

Illustration 31.If temperature of ideal black body is increased by 50%, what will be percentage increase in quantity of radiations

emitted from its surface.

Solution

E µ T4 and \ E' µ (1.5)4T4 µ 4 4

4 4 415 3 81T T T

10 2 16é ù é ùµ µê ú ê úë û ë û

E ' E100%

E-

´ =

4 4

4

81T T

16 100%T

é ù-ê ú´ê ú

ê úê úë û

= 406 % » 400 %

Illustration 32.

If temperature of ideal black body is decreased from T to T2

than find out percentage loss in emissive rate

Solution

E µ T4, 4 4T T

E '2 16

é ùµ ê úë û

E E 1 15100% 1 100% 100%

E 16 16

¢-é ù é ù´ = - ´ = ´ê ú ê úë û ë û » 94% Remaining is 6% ( Approx.)

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

32 E


Calculate the temperature at which a perfect black body radiates at the rate of 5.67 W cm–2. Stefan's constant

is 5.67 × 10–8 J s–1 m–2 K–4.

Solution

Given E = 5.67 W cm–2 = 5.67 × 10+4 W m–2 , s = 5.67 × 10–8 J s–1 m–2 K–4

Using,E = sT4; T4 = Es

or T = Es

LNM

OQP

14

=

14 4

8

5.67 10

5.67 10

+

-

é ù´ê ú

´ê úë û = (1012)1/4 = 1000 K

Illustration 34.The temperature of furnace is 2000°C, in its spectrum the maximum intensity is obtained at about 4000Å,

If the maximum intensity is at 2000Å calculate the temperature of the furnace in °C.

Solutionby using lmT = b, 4000 (2000+273) = 2000(T) Þ T = 4546K

The temperature of furnace = 4546 – 273 = 4273 °C

Illustration 35.Two bodies A and B have thermal emissiviities of 0.01 and 0.81 respectively. The outer surface areas of

the two bodies are same, the two bodies emit total radiant power at the same rate. The wavelength lB

corresponding to maximum spectral radiancy of B is shifted from the wavelength corresponding to maximum

spectral radiancy in the radiation of A by 1.0 mm. If the temperature of A is 5802K, Calculate :-

(a) The temperature of B

(b) Wavelength lB

Solution

(a) As both bodies A and B having same radiant power

\ PA = PB Þ eAsAATA4 = eBsABTB

4 Þ (0.01)sATA4 = (0.81)sATB

4

1/ 4A

B A0.01 T 5802

T T 1934 K0.81 3 3

æ ö= = = =ç ÷è ø

(b) According to wein's displacement law

lATA = lBTB Þ lB = 58021934

æ öç ÷è ø

lA = 3lA

As lB – lA = 1 µm Þ lB – B

3l

= 1 µm Þ B23l

= 1 µm Þ lB = 1.5 µm .

Illustration 36.

When a metallic body is heated in a furnace, then what colour will appear as temperature increases.

Solution. µlm

1T

As Temperature increases colour of body will appear from red, yellow, green, blue and then white.Illustration 37.

Define (i) Steady state and (ii) Temperature gradient in conduction of heat through a conducting rod.Solution

(i) When one end of a rod is heated, the temperature of various points of the rod changes continuouslybut after some time a state is reached, when the temperature of each cross–section becomes steadywhich is called steady state. In this state the heat received by any section will be totally transfered tothe next section so no heat is absorbed by any cross section.

(ii) Temperature gradient is defined as the rate of change of temperature with distance in the directionof flow of heat.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

33E


Assuming Newton's law of cooling to be valid. The temperature of body changes from 60°C to 40°Cin 7 minutes. Temperature of surroundings being 10°C, Find its temperature after next 7 minutes.

Solution

According to Newton's law of cooling 1 2T – Tt

= K 1 2

0T T

– T2+æ ö

ç ÷è ø

Since the temperature decreases from 60°C to 40°C in 7 minutes

60 407–

= K 60 40

210

+FHG

IKJ– Þ

207

= K (50 – 10) Þ K = 1

14

If the temperature of object becoms T2 in next 7 minutes then 240 – T

7 =

114

2T40– 10

2+æ ö

ç ÷è ø

Þ 40 – T2 = 14

(40 + T2 – 20) Þ 160 – 4T2 = 20 + T2

Þ 5T2 = 140 Þ T2 = 28°C

BEGINNER'S BOX-4

1. Explain why :

(a) a body with large reflectivity is a poor emitter.

(b) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives

too low value for the temperature of a red hot iron piece in the open, but gives a correct value for

the temperature when the same piece is in the furnace

(c) the earth without its atmosphere would be inhospitably cold

(d) Heat is generated continuously in an electric heater but its temperature remains constant after some time.

(e) The bottom of a cooking vessel is made black ?

(f) On winter night you feel warmer when clouds cover the sky than when the sky is clear.

(g) A thermos or vacuum flask can keep hot things hot and cold things cold for a long time, how?

2. Two spherical ideal black bodies of radii r1 and r2 are having surface temperature T1 and T2 respectively,

If both radiate the same power. Then calculate the ratio of TT

1

2.

3. If a liquid takes 30 sec. in cooling from 80°C to 70°C and 70 sec in cooling from 60°C to 50°C, then find

the room temperature.

4. A body cools in 10 minutes from 60°C to 40°C. What will be its temperature after next 10 minutes? The

temperature of the surrounding is 10°C.

5. Calculate the temperature of the black body from given graph.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

34 E

Pre-Medical : Physics ALLENKINETIC THEORY OF GASES

The properties of the gases are entirely different from those of solid and liquid. In case of gases, thermal

expansion is very large as compared to solids and liquids .To state the conditions of a gas, its volume, pressure

and temperature must be specified.

Intermolecular force Solid > liquid > real gas > ideal gas (zero)

Potential energy Solid < liquid < real gas < ideal gas (zero)

At a given temperature for solid, liquid and gas:

(i) Internal kinetic energy : Same for all

(ii) Internal potential Energy : Maximum for ideal gas (PE = 0) and Minimum for solids (PE = –ve)

(iii) Internal Energy : Maximum for Ideal gas and Minimum for solid

Notation :

m = Molar amount of gas = M/Mw = N/N0 m = mass of a gas molecule

N0 = Avogadro constant = 6.023 × 1023 molecules/mole M = mass of gas

R = Universal gas constant = 8.31 J/mole– k Mw = molecular weight

» 2 cal/mole-k, [Dimension = ML2T-2m-1q-1] N = No. of gas molecules

r = Specific gas constant (r = R/Mw) n = Molecular density (n = N/V)

k = Boltzmann constant = (k = R/N0) = 1.38 x 10–23 J/kelvin r = gas density (r = M/V = mn)

N.T.P. S.T.P.

(Normal temperature and pressure) (Standard Temperature and Pressure)

Temperature 0° C = 273.15 K 0.01° C = 273.16K

Pressure 1 atm = 1.01325 × 105 N/m2 1 atm

= 1.01325 × 105 pascal

Volume 22.4 litre 22.4 litre

1 lit = 10–3 m3 = 103 cm3 = 103cc = 103ml

rwater=1000 kg/m3 = 1 g/cc, rice=900 kg/m3 = 0.9 g/cc

Ideal Gas Concept

• Volume of gas molecules is negligible as compared to volume of container.

So volume of gas = volume of container (Except 0 K)

• No intermoleculer forces act between gas molecules.

Properties of Ideal Gas

• A gas which follows all gas laws and gas equation at every possible temperature and pressure is known as

ideal or perfect gas.

• Ideal gas molecules can do only translational motion, so their kinetic energy is only translational kinetic energy

• Ideal gas can not be liquified because IMF is zero

• Potential energy of ideal gas is zero so internal energy of ideal gas is perfectly translational K.E. of gas.

It is directly proportional to absolute temperature.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

35E

Pre-Medical : PhysicsALLENSo, internal energy depends only and only on its temperature.

Etrans µ T

For a substance U = UKE + UPE

UKE : depends only on T, UPE : depends upon intermolecular forces (Always negative)

• Specific heat of ideal gas is constant quantity and it does not change with temperature

• All real gases behaves as an ideal gas at high temperature and low pressure and low density.

• Gas molecules have point mass and negligible volume and velocity is very high (107 cm/s). That's why there

is no effect of gravity on them.

Equation of state for Ideal gas

PV µRT= Þ w

MPV RT

M= =

0

mNRT

mN

é ùê úë û

=0

RN

é ùê úë û

N T = NkT where µ = number of moles of gas

k = Boltzamann constant

Þ w

P RT kTM m

= =r

Gas Laws• Boyle's Law

According to it for a given mass of an ideal gas at constant temperature, the volume of a gas is inversely

proportional to its pressure, i.e., µ1

VP

if m and T = Constant Þ 1 1 2 2P V P V=

m=constantT=constant

P

V

PV

V

PV

P

• Charle's LawAccording to it for a given mass of an ideal gas at constant pressure, volume of a gas is directly proportional

to its absolute temperature, i.e. V µ T if m and P = Constant Þ 1 1

2 2

V TV T

=

V

T(k)

VT

V

VT

T(k)

Gay–Lussac's LawAccording to it, for a given mass of an ideal gas at constant volume, pressure of a gas is directly proportional

to its absolute temperature, i.e., P µ T if m and V = constant Þ 1 1

2 2

P TP T

=

P

T(k)

PT

P

PT

T(k)

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

36 E

Pre-Medical : Physics ALLENAvogadro's Law

According to it, at same temperature and pressure, equal volume of all gases contain equal number of molecules,

i.e., 1 2N = N if P,V and T are same.

Dalton's Partial Pressure Mixture Law :

According to it, the pressure exerted by mixture of non-reactive gases is equal to the sum of partial pressure

of each component gases present in the mixture, ie., P = P1 + P2 + ....


By increasing temperature of gas by 5° C its pressure increases by 0.5% from its initial value at constant

volume then what is initial temperature of gas ?

Solution

Q At constant volume T µ P \D DTT

PP

´ = ´ =100 100 05. Þ5 100

T 1000K0.5´

= =

Illustration 40.

Calculate the value of universal gas constant at STP.

Solution

Universal gas constant is given by RPVT

=

One mole of all gases at S.T.P. occupy volume V = 22·4 litre = 22·4 × 10–3 m3

P = 760 mm of Hg = 760 × 10–3 × 13·6 × 103 × 9.80 N m–2, T = 273.16 K

\ R = 3 3 3760 10 13.6 10 9.80 22.4 10

273.16

- -´ ´ ´ ´ ´ ´= 8·314 J mol–1 K–1

Illustration 41.

A closed container of volume 0.02 m3 contains a mixture of neon and argon gases at a temperature of 27°C

and pressure of 1 × 105 N/m2. The total mass of the mixture is 28 g. If the gram molecular weights of

neon and argon are 20 and 40 respectively, find the mass of the individual gases in the container, assuming

them to be ideal. Given : R = 8.314 J/mol-K.

Solution

Let m gram be the mass of neon. Then, the mass of argon is (28 – m)g.

Total number of moles of the mixture,m 28 m 28 m

µ20 40 40

- += + = ...(i)

Now, 5PV 1 10 0.02

0.8RT 8.314 300

´ ´m = = =

´...(ii)

By (i) and (ii),+

=28 m

0.840

Þ 28 + m = 32 Þ m = 4 gram

or mass of argon = (28 – 4)g = 24 g

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

37E


Calculate the temperature of the Sun if density is 1.4 g cm–3, pressure is 1.4 × 109 atmosphere and average

molecular weight of gases in the Sun in 2 g/mole. [Given R = 8.4 J mol–1K–1]

Solution

PV = µRT Þ PV

TR

=m ...(i) But

w

MM

m = and r =MV

\ w

VMr

m =

From equation (i)w wP V M PM

TV R R

= =r r =

9 5 31.4 10 1.01 10 2 101.4 1000 8.4

-´ ´ ´ ´ ´´ ´

= 2.4 × 107 K

Illustration 43.

At the top of a mountain a thermometer reads 7°C and barometer reads 70 cm of Hg. At the bottom of

the mountain they read 27°C and 76 cm of Hg respectively. Compare the density of the air at the top with

that at the bottom.

Solution

By gas equation =w

MPV RT

MÞ =

r w

P RT M w

M M and

M V

é ùm = = rê ú

ë ûQ

Now as MW and R are same for top and bottom T B

P PT T

é ù é ù=ê ú ê úr rë û ë û

So T T B

B B T

P TP T

r= ´

r70 300 7576 280 76

= ´ = = 0.9868

Illustration 44.

A sample of oxygen of volume of 500 cc at a pressure of 2 atm is compressed to a volume of

400 cc. What pressure is needed to do this if the temperature is kept constant ?

Solution

Temperature is constant, so P1 V1 = P2V2 \ é ù= = ê úë û

12 1

2

V 500P P 2

V 400 = 2.5 atm

Illustration 45.

A vessel of volume 8.0 × 10–3 m3 contains an ideal gas at 300 K and 200 k Pa. The gas is allowed to

leak till the pressure falls to 125 kPa. Calculate the amount of the gas leaked assuming that the temperature

remains constant.

Solution

As the gas leaks out, the volume and the temperature of the remaining gas do not change. The number of

moles of the gas in the vessel in given by µ = PVRT

.

The number of moles in the vessel before the leakage is 11

P VRT

m = and that after the leakage is 22

P Vµ

RT= .

The amount leaked is µ1 – µ2 = -1 2(P P )VRT

-- ´ ´ ´=

´

3 3(200 125) 10 8.0 108.3 300

= 0.24 mole

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

38 E


1500 ml of a gas at a room temperature of 23°C is inhaled by a person whose body temperature is 37°C,

if the pressure and mass stay constant, what will be the volume of the gas in the lungs of the person ?

Solution

T1 = 273 + 23 = 296 K; T2 = 273 + 37 = 310 K. Pressure and amount of the gas are kept constant,

So =1 2

1 2

V VT T \ = ´ 2

2 11

TV V

T = 310

1500296

´ = 1570.95 ml

Illustration 47.

A sample of O2 is at a pressure of 1 atm when the volume is 100 ml and its temperature is 27°C. What

will be the temperature of the gas if the pressure becomes 2 atm and volume remains 100 ml.

Solution

T1 = 273 + 27 = 300 K

For constant volume =1 2

1 2

P PT T Þ = ´ 2

2 11

PT T

P= ´

2300

1 = 600 K = 600 – 273 = 327°C

BEGINNER'S BOX-5

1. A vessel of volume 8.3 × 10–3 m3 contains an ideal gas at temperature 27°C and pressure 200 kPa. The

gas is allowed to leak till the pressure falls to 100 kPa and temperature increases to 327°C. What is the amount

of gas in moles will be leaked out ?

2. Two closed vessels of equal volume contain air at 105 kPa, 300 K and are connected through a narrow tube.

If one of the vessels is now maintained at 300 K and other at 400 K, What will be the pressure in the vessels ?

3. The volume of a gas is 1 litre at the pressure 1.2 × 107 Nm–2 and temperature 400 K. Calculate the number

of molecules in the gas.

4. A vessel contains two non- reactive gases : neon (monatomic) and oxygen (diatomic) The ratio of their partial

pressures is 3:2. Estimate the ratio of

(a) number of molecules and

(b) mass density of neon and oxygen in the vessel.

Atomic mass of Ne = 20.2 u. molecular mass of O2 = 32.0 u.

The kinetic theory of gases

• Rudolph Claussius (1822–88) and James Clark Maxwell (1831–75) developed the kinetic theory of

gases in order to explain gas laws in terms of the motion of the gas molecules. The theory is based

on following assumptions regarding the motion of molecules and the nature of the gases.

Basic postulates of Kinetic theory of gases

• Every gas consists of extremely small particles known as molecules. The molecules of a given gas are

all identical but are different than those of another gas.

• The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.

• The size is negligible in comparision to inter molecular distance (10–9 m)

Assumptions regarding motion :

• Molecules of a gas keep on moving randomly in all possible direction with all possible velocities.

• The speed of gas molecules lie between zero and infinity (very high speed).

• The number of molecules moving with most probable speed is maximum.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

39E

Pre-Medical : PhysicsALLENAssumptions regarding collision:

• The gas molecules keep colliding among themselves as well as with the walls of containing vessel. These

collision are perfectly elastic. (ie., the total energy before collision = total energy after the collision.)

Assumptions regarding force:

• No attractive or repulsive force acts between gas molecules.

• Gravitational attraction among the molecules is ineffective due to extremely small masses and very high

speed of molecules.

Assumptions regarding pressure:

• Molecules constantly collide with the walls of container due to which their momentum changes. This

change in momentum is transferred to the walls of the container. Consequently pressure is exerted

by gas molecules on the walls of container.

Assumptions regarding density:

• The density of gas is constant at all points of the container.

Properties/Assumptions of Ideal gas

• The molecules of a gas are in a state of continuous random motion. They move with all possible velocities

in all possible directions. They obey Newton's law of motion.

• Mean momentum = 0; Mean velocity < rv > = 0; < v2 > ¹ 0 (Non zero); < v3 > = < v5 > = 0

• The average distance travelled by a molecule between two successive collisions is called as mean free

path (lm) of the molecule.

• The time during which a collision takes place is negligible as compared to time taken by the molecule to

cover the mean free path. At NTP ratio of time of collision to time of motion to cover mean free path

is 10–8 : 1.

• When a gas is taken into a vessel it is uniformly distributed in entire volume of vessel such that its

mass density, moleculer density, motion of molecules etc. all are identical for all direction, therefore

root mean velocity

= =2 2 2x y zv v v are equal. Pressure exerted by the gas in all direction Px = Py = Pz = P ® equal

• All those assumptions can be justified, if number of gas molecules are taken very large

i.e., 1023 molecules/cm3.

Different speeds of gas molecules

• Average velocity

Because molecules are in random motion in all possible direction with all possible velocity. Therefore, the

average velocity of the gas molecules in container is zero. < rv > =

r r rv v v

NN1 2+ +............

= 0

RMS speed of molecules vrms= r3P

= w

3RTM =

3kTm

= 1.73kTm

Average speed or Mean speed of molecules :

By maxwell’s velocity distribution law vM or <||rv > º vmean

<||rv > = vmean =

+ +1 2 n|v | |v | ..........|v |N

r r r

=p r8 P

= 8RTMwp

= 8kT

mp= 1.59

kTm

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

40 E

Pre-Medical : Physics ALLENMost probable speed of molecules (vmp)At a given temperature, the speed to which maximum number of molecules belongs is called as most probable

speed (vmp) vmp = r2P

= 2RTMw

= 2kTm

= 1.41 kTm

Velocity of sound in gas medium (vs ) vsound = grP

= w

RTMg

= gkTm

• At any temperature vrms > vMean > vMP > vsound (always)

• For a gas at any temperature (T)v

vrms

sound=

3g ,

vv

rms

MP =

32

• A temperature is not possible at which above order can be changed

vrms ¹ vMean ¹ vMP ¹ vsound (always )

Maxwell's law of distribution of velocities

v

vmp (T )1 vmp vmp(T )2 (T )3

vmp= most probable speed

Nmax= maximum number of molecules at given temperature

num

berof

mol

ecul

es(N

)

velocity of molecule

Nmax(T )1

Nmax

Nmax

T =500K1

T =1000K2

T =2000K3

(T )2

(T )3

T3 > T2 > T1

Expression for Pressure of An Ideal gas

Consider an ideal gas enclosed in a cubical vessel of length l.

z

y

x

l

l0

A

B

D

C

Suppose there are 'N' molecules in a gas which are moving with

velocities 1 2 Nv , v ........vr r r

.

If we consider any single molecule then its instantaneous velocity vr can be expressed as x y z

ˆˆ ˆv v i v j v k= + +r

Due to random motion of the molecule vx = vy = vz ; x y zv v 3 v 3 v 3= = =r

= + +2 2 2x y zv v v

Suppose a molecule of mass m is moving with a velocity vx towards the face ABCD. It strikes the face ofthe cubical vessel and returns back to strike the opposite face.

Change in momentum of the molecule per collision Dp = – mvx – mvx = – 2 mvx

Momentum transferred to the wall of the vessel per molecule per collision Dp= 2 mvx

The distance travelled by the molecule in going to face ABCD and coming back is 2l.

So, the time between two successive collision is D =x

2t

vl

Number of collision per sec per molecule is = xc

vf

2l = molecule velocity

mean free path; =

lrms

cm

vf or =

lm

cm

vf

Hence momentum transferred to the wall per second by the molecule is equal to force applied on the wall

therefore, force F = (2 mvx) xv

2l=

2 2x

x

mv mvAs V 3V

3é ù= =ë û

l l

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

41E


Pressure exerted by gas molecule =F

PA

= ´

21 mv3 Al

Þ P= 21 mv

3 V [ ]´ =A VQ l

pressure exerted by gas P = å21 mv

3 V=

21 mv N3 V N

´å =å 2v1 mN

3 V N= 2

rms

1 mNv

3 V

2rms

w

3PV 3 RTv

M µMm

= = Þ =rmsw

3RTv

M , P = = r2 2rms rms

1 M 1v v

3 V 3

• Average number of molecules for each wall = N6

.

No. of molecules along each axis = N3

(Nx = Ny = Nz)

•2

2 2 2 rmsx y z

vv v v

3= = = Root mean square velocity along any axis for gas molecule is (vrms)c= (vrms)y= (vrms)z=

vrms

3

All gas laws and gas equation can be obtained by expression of pressure of gas (except Joule’s law)

Degree of freedom (f)

• The number of independent ways in which a molecule or an atom can exhibit motion or have energy is calledits degrees of freedom.

• The number of independent coordinates required to specify the dynamic state of a system is called its degreesof freedom.

• The degrees of freedom are of three types :

(a) Translational Degree of freedom : There are maximum three degree of freedom correspondingto translational motion.

(b) Rotational Degree of freedom : The number of degrees of freedom in this case depends on thestructure of the molecule.

(c) Vibrational Degree of freedom : It is exhibited at high temperature.

Degrees of freedom for different gases according to atomicity of gas at low temperature

Atomicity of gas Translational Rotational Total

Monoatomic

Ex. Ar, Ne, Ideal gas etc 3 0 3

y

zx

Diatomic

Ex. O2, Cl2, N2 etc. 3 2 5

Triatomic (linear) Ex. CO

2, C

2H2

3 2 5 O=C=O

Triatomic (Non–linear)

or Polyatomic Ex. H2O, NH3, CH4

3 3 6

At high temperature a diatomic molecule has 7 degrees of freedom. (3 translational, 2 rotational and 2 vibrational)

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

42 E

Pre-Medical : Physics ALLENMaxwell's law of equipartition of energy

The total kinetic energy of a gas molecules is equally distributed among its all degree of freedom and the

energy associated with each degree of freedom at absolute temperature T is 1

kT2

For one molecule of gas

Energy related with each degree of freedom = 1

kT2

Energy related with all degree of freedom = f

kT2

Q

22 2 2 rmsx y z

vv v v

3= = = Þ

12

32

2mv kTrms =

So energy related with one degree of freedom = = =2rms1 v 3 kT 1

m kT2 3 2 3 2

Different K.E. of gas (Internal Energy)

• Translational kinetic energy (ET) ET = 2rms

1Mv

2 =

3PV

2

Kinetic energy of volume V is = 2rms

1Mv

2 [Note : Total internal energy of ideal gas is total kinetic energy]

• Energy per unit volume or energy density (EV) V

Total energy EE

Volume V= = ;

é ù= ê úë û2

V rms

1 ME v

2 V 2rms

1v

2= r

Q é ù= rê úë û

2rms

2 1P v

3 2 \ =V

3E P

2

• Molar K.E. or Mean Molar K.E. (E) = 2w rms

1E M v

2 for N0 molecules or Mw (gram) ;

= = 0

3 3E RT N kT

2 2

• Molecular kinetic energy or mean molecular K.E. (E ) = 2w rms

1E M v

2, = = =

0 0

E 3 RT 3E kT

N 2 N 2

• Kinetic energy of 1 g mass (Em) = 0

W 0

N KT3 RT 3 3 KT2 M 2 N m 2 m

´= =

SPECIFIC HEAT CAPACITYMonoatomic GasesThe molecule of a monoatomic gas has only three translational degrees of freedom. Thus, the average energyof a molecule at temperature T is (3/2)kT.The total internal energy of a mole of such a gas is

U = A3 3

kT N RT2 2

´ =

The molar specific heat at constant volume, Cv, is

Cv (monoatomic gas) = dU 3

RdT 2

=

Diatomic GasesAs explained earlier, a diatomic molecule treated as a rigid rotator like a dumbbell has 5 degrees of freedom: 3 translational and 2 rotational. Using the law of equipartition of energy, the total internal energy of a moleof such a gas is

U = 5

kT2

× NA = 5

RT2

The molar specific heats are then given by

Cv (rigid diatomic) = 5

R2

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

43E

Pre-Medical : PhysicsALLENMEAN FREE PATH

A molecule in its path undergoes a number of collisions so the path traversed by it is not a straight line

but somewhat zigzag. Between two successive collisions a molecule travels in a straight line with uniform

velocity. As motion is random thus the distance travelled by molecule between two succesive collisions is

not always equal.

The average distance travelled by a molecule between two succesive collisions is called as mean free path

(lm) of the molecule.

m 2

1

2 d nl =

p

Here n = NV

= No. of molecules per unit volume

d = diameter of a molecule

GOLDEN KEY POINTS

• At a given temperature graph drawn between molecular velocity and number of molecules is known as velocity

distribution curve.

• The velocities of molecules of a gas are in between zero and infinity (0 – ¥ )

• With the increase in the temperature, the most probable velocity and maximum molecule velocity both

increases.

• The number of molecules within certain velocity range is constant although the velocity of molecule changes

continuously at particular temperature.

• The area enclosed between the (N – v) curve and the velocity axis represents the total number of molecules.

On the basis of velocity distribution Maxwell established the law of equipartition of energy for gases at any

temperature.

• Except 0 K, at any temperature T , E > Em > E

• At a common temperature, for all ideal gas

E and E are same while Em is different and depends upon nature of gas (Mw or m)

• For thermal equilibrium of gases, temperature of each gas is same and this temperature called as temperature

of mixture (Tm) which can be find out on basis of conservation of energy (All gases are of same atomicity).

+ + += =

+åå

1 1 2 2 n nm

1 2 n

NT N T N T ......... N TT

N N ......NN

• 1 mole gas : Mean translational kinetic energy = 3

RT2

; Total kinetic energy = f

RT2

1 molecule of gases : Mean translational kinetic energy = 3

kT2

; Total kinetic energy = f

kT2 ;

¦ ® Degree of freedom

• Specific heats of all substances approach zero as T ® 0. This is related to the fact that degrees of freedom

get frozen and ineffective at low temperatures.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

44 E

Pre-Medical : Physics ALLENIllustrations

Illustration 48.

The velocities of ten particles in ms–1 are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate

(i) average speed and (ii) rms speed (iii) most probable speed.

Solution

(i) average speed, vav= 0 2 3 4 4 4 5 5 6 9

10+ + + + + + + + +

= 4210

= 4·2 ms–1

(ii) rms speed, vrms =

1 22 2 2 2 2 2 2 2 2 2(0) (2) (3) (4) (4) (4) (5) (5) (6) (9)10

é ù+ + + + + + + + +ê úë û

= 1/2

22810

é ùê úë û

= 4.77 ms–1

(iii) most probable speed vmp = 4 m/s

Illustration 49.

At what temperature root mean square velocity of hydrogen becomes double of its value at S.T.P., pressure

remaining constant ?

Solution

Let v1 be the r.m.s. velocity at S.T.P. and v2 be the r.m.s. velocity at unknown temperature T2.

\v

v

TT

12

22

1

2=

or T2 = T1

2

2

1

vv

é ùê úë û

= 273 × (2)2 = 273 × 4 = 1092 K = (1092 – 273) = 819°C

Illustration 50.

Calculate rms velocity of oxygen molecule at 27°C

Solution

Temperature, T = 27° C

Þ 273 + 27 = 300 K,

Molecular weight of oxygen = 32 × 10–3 kg and R = 8·31 J mol–1 K–1

rms velocity is vrms = 3RTM

= 3 8 31 300

32 10 3

´ × ´´ - = 483·5 ms–1

Illustration 51.

Calculate the kinetic energy of a gram moelcule of argon at 127°C.

Solution

Temperature, T = 127°C = 273 + 127 = 400 K, R = 8.31 J/mol K

K.E. per gram molecule of argon = 32

R T = 32

× 8.31 × 400 = 4986 J

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

45E


The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 molecules are colliding per second on a stationary

wall of area 2 cm2 at an angle of 45° to the normal to the wall and reflected elastically with a speed 103

m/s. Find the pressure exerted on the wall. (in N/m2)

Solution

As the impact is elastic \ = = =1 2p p p mvr r

= 3.32 × 10–24 kg m/s

45°45°

A

B

O

P1

P2

The change in momentum along the normal 2 1p p p 2pcos45 2pD = - = ° =r r

If f is the collision frequency then force applied on the wall D

= = D ´ =Dp

F p f 2pft

\ Pressure = =F 2pf

PA A

= 24 23

4

2 3.32 10 102 10

-

-

´ ´ ´´

= ´ 3 22.347 10 N /m

BEGINNER'S BOX-61. Given : Avogadro number N = 6.02 × 1023 molecules/mole and Boltzmann's constant

k = 1.38 × 10–23 J/(molecule-K).

Calculate %

(a) The average kinetic energy of translation of an oxygen molecule at 27°C

(b) The total kinetic energy of an oxygen molecule at 27°C

(c) The total kinetic energy of 1 mole of oxygen gas at 27°C.

2. The temperature of gas is –73°C. To what temperature should it be heated so that

(a) average kinetic energy of the molecule is doubled ?

(b) the root mean square velocity of the molecules is doubled ?

3. For a gas R

CP= 0.4 . For this gas calculate the following –

(a) atomicity and degree of freedom

(b) value of Cv and g

(c) mean gram - molecular kinetic energy at 300 K temperature

4. If three molecules are having speeds v1, v2 and v3 respectively, then what will be their average speed and rootmean square speed ?

5. Four molecules of a gas are having speeds of 1, 4, 8 and 16 ms–1. Find the root mean square velocity ofthe gas molecules ?

6. A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixture is 27°C.Obtain the ratio of

(a) Average translational kinetic energy per molecule, and

(b) root mean square speed (nrms ) of the molecules of the two gases.

Atomic mass of argon = 39.9 u :

Molecular mass of chlorine = 70.9 u.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

46 E

Pre-Medical : Physics ALLENTHERMODYNAMICS

Branch of physics which deals with the inter–conversion between heat energy and any other form of energyis known as thermodynamics. In this branch of physics we deal with the processes involving heat, work andinternal energy. In this branch of science the conversion of heat into mechanical work and vice versa is studied.

l Thermodynamic System

The system which can be represented in terms of pressure (P), volume (V) and temperature (T), is knownthermodynamic system. A specified portion of matter consisting of one or more substances on which theeffects of thermodynamic variables such as temperature, volume and pressure are to be studied, is called asystem.

e.g. A gas enclosed in a cylinder fitted with a piston is a system.

l Surrounding

Anything outside the system, which exchanges energy with the system and which tends to change the propertiesof the system is called its surrounding.

l Heterogeneous System

A system which is not uniform throughout is said to be heterogeneous. e.g. A system consisting of two ormore immiscible liquids.

l Homogeneous System

A system is said to be homogeneous if it is completely uniform throughout. e.g. Pure solid or liquid.

l Isolated System

A system in which no exchange of matter and energy with the surrounding take place, is said to be an isolatedsystem.

l Universe

The system and its surrounding are together known as the universe.

l Thermodynamic variables of the system

(i) Composition (m) (ii) Temperature (T) (iii) Volume (V) (iv) Pressure (P) (v) Mass

l Thermodynamic state

The state of a system can be described completely by composition, temperature, volume and pressure.

If a system is homogeneous and has definite mass and composition, then the state of the system can be describedby the remaining three variables namely temperature, pressure and volume. These variables are inter relatedby equation PV = µRT. The thermodynamic state of the system is its condition as identified by two independentthermodynamic variables (P, V or P, T or V, T).

l Internal Energy

Internal energy of a system is the energy possessed by the system due to molecular motion and molecularconfiguration. The energy due to molecular motion is called internal kinetic energy (Uk) and that due to molecularconfiguration is called internal potential energy (Up). dU = dUk + dUp

If there no intermolecular forces, then dUp = 0 and dU = dUk = m cv dT [ideal gas]

cv = Specific heat at constant volume and dT = Infinitesimal change in temperature

m = Mass of system M = Molecular weight

Molar heat capacity Cv = Mcv For µ–moles of ideal gas v v

mdU µC dT C dT

M= =

Internal energy in the absence of inter–molecular forces is simply the function of temperature and state only,it is independent of path followed. DU = Uf – Ui

Ui = Internal energies in initial state and Uf = Internal energies in final state

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

47E

Pre-Medical : PhysicsALLENl Thermodynamic Processes

In the thermodynamic process pressure, volume, temperature and entropy of the system change with time.

Thermodynamic process is said to take place if change occurs in the state of a thermodynamic system.

l Sign convention used for the study of thermodynamic processes

Heat gained by a system Positive

Heat lost by a system Negative

Work done by a system Positive

Work done on the system Negative

Increase in the internal energy of system Positive

Decrease in the internal energy of system Negative

l Indicator Diagram or P–V Diagram

In the equation of state of a gas PV = µRT two thermodynamic variables are sufficient to describe the behaviorof a thermodynamic system.

If any two of the three variables P, V and T are known then the third can be calculated.

P–V diagram is a graph between the volume V and the pressure P of the system.

The volume is plotted against X–axis while the pressure is plotted against Y–axis.

The point A represents the initial stage of the system. Initial pressure of the system is Pi and initial volumeof the system Vi .

The point B represents the final state of the system. Pf and Vf are the final(P , V )i i

(P , V )f f

P

V

A

Bpressure and final volume respectively of the system. The points between A andB represent the intermediate states of the system. With the help of the indicatordiagram we calculate the amount of work done by the gas or on the gas duringexpansion or compression.

l Cyclic process

Cyclic process is a thermodynamic process in which the system returns to its initial stage after undergoinga series of changes.

l Non–cyclic process

Non–cyclic process is a process in which the system does not return to its initial stage.

l Quasi–static or equilibrium process

Quasi–static is a thermodynamic process which proceeds extremely slowly such that at every instant of time,the temperature and pressure are the same in all parts of the system.

l Reversible and Irreversible processes

A reversible process is one in which the changes in heat and work of direct process from initial to a finalstate are exactly retraced in opposite sense in the reverse process and the system and surroundings are leftin their initial states. The reversibility is an ideal concept and can not be realized in practice.

The process which is not reversible is the irreversible process. In nature the processes are irreversible.

Work done by thermodynamic system

One of the simple example of a thermodynamic system dx

F = P× A

(P, V)area of piston is A

is a gas in a cylinder with a movable piston.

• If the gas expands against the pistonGas exerts a force on the piston and displace it

through a distance and does work on the piston.

• If the piston compresses the gasWhen piston moved inward, work is done on the gas.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

48 E

Pre-Medical : Physics ALLEN• The work associated with volume changes

If pressure of gas on the piston = P.

Then the force on the piston due to gas is F = PA

When the piston is pushed outward an infinitesimal distance dx,

(P , V )i i

(P , V )f f

P

V

A

B

dV

P

the work done by the gas is dW = F × dx = PA dx

The change in volume of the gas is dV = Adx, \ dW = PdV

For a finite change in volume from Vi to Vf, this equation is then integrated between Vi to Vf to find the

net work done f f

i i

V V

V VW dW PdV= =ò ò

Hence the work done by a gas is equal to the area under P–V graph.

Following different cases are possible.

(i) Volume is constant

or

V = constant and W = 0

(ii) Volume is increasing

P

V

FinalInitialA B

or

P

V

Final

InitialA

B

V is increasing WAB > 0 WAB = Shaded area

(iii) Volume is decreasing

P

V

Initial

Final

B

A

or

P

V

Initial FinalB A

V is decreasing WAB < 0 WAB = – Shaded area

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

49E

Pre-Medical : PhysicsALLEN(iv) Cyclic process

P

V

P

V

Wclockwise cycle = + Shaded area Wanticlockwise cycle = – Shaded area

Work Done in Clockwise Cycle

P

V

B

A =

P

V

B

A+

P

V

B

A

WAB (Positive) WBA (Negative)

Wcyclic = WAB (positive) + WAB (negative) = area o

f

closed

path

First law of thermodynamics

Let a gas in a cylinder with a movable piston changes from an initial

equilibrium state to a final equilibrium state.

System change its state through path 'a' :

The heat absorbed by the system in this process = dQa

P

V

a

b

c

initial

final

The work done by the system = dWa

Again for path 'b' :

Heat absorbed by the system = dQb , Work done by the system = dWb.

It is experimental fact that the dQa – dWa = dQb – dWb

Both dQ and dW depend on the thermodynamic path taken between two equilibrium states, but difference

(dQ – dW) does not depends on path in between two definite states of the system.

So, there is a function (internal energy) of the thermodynamic coordinates (P, V and T) whose final value

(Uf) minus its initial value (Ui) equals the change dQ – dW in the process.

dU = dQ – dW. This is the first law of thermodynamics.

Heat supplied to the system and work done by the system are path dependent so they are denoted by dQ

and dW respectively. Change in internal energy DU = Uf – Ui does not depends on path it depends only

on initial and final positions of the system. So, it is denoted by dU (or DU)

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

50 E

Pre-Medical : Physics ALLENFirst Law of Thermodynamics

If some quantity of heat is supplied to a system capable of doing external work, then the quantity of heat

absorbed by the system is equal to the sum of the increase in the internal energy of the system and the external

work done by the system. Q = dU + Wd d or Q = W + UD

• This law is applicable to every process in nature

• The first law of thermodynamics introduces the concept of internal energy.

• The first law of thermodynamics is based on the law of conservation of energy.

• dQ, dU and dW must be expressed in the same units (either in units of work or in units of heat).

• This law is applicable to all the three phases of matter, i.e., solid, liquid and gas.

• dU is a characteristic of the state of a system, it may be any type of internal energy–translational kinetic

energy, vibrational, rotational kinetic energy, binding energy etc.

Application of First Law of Thermodynamics

• Melting Process :

When a substance melts, the change in volume (dV) is very small and can, therefore, be neglected. The temperature

of a substance remains unchanged during melting process.

Let us consider the melting of a mass m of the solid. Let L be the latent heat of fusion i.e., the heat required

L to change a unit mass of a solid to liquid phase at constant temperature.

Heat absorbed during melting process, Q = mL

By the first law of thermodynamics, Q=DU+W=DU+PDV, where DV is volume change during phase change.

• Boiling Process :

When a liquid is heated, it changes into vapour at constant temperature (called boiling point) and pressure.

When water is heated at normal atmospheric pressure, it boils at 100°C. The temperature remains unchanged

during the boiling process. Let us consider the vaporisation of liquid of mass m. Let Vl and Vv be the volumes

of the liquid and vapours respectively.

The work done in expanding at constant temperature and pressure P, dW = PDV = P (Vv – Vl)

Let the latent heat of vaporisation = L \ Heat absorbed during boiling process, dQ = mL

Let Ul and Uv be the internal energies of the liquid and vapours respectively then DU = Uv – U

l

According to the first law of thermodynamics, dQ = DU + dW \ mL = (Uv – Ul) + P(Vv – V

l)


The pressure of one mole monoatomic gas increases linearly from 4 × 105 Nm–2 to 8 × 10+5 Nm–2 when

its volume increases from 0.2m3 to 0.5 m3. Calculate.

(i) Work done by the gas, (ii) Increase in the internal energy,

(iii) Amount of heat supplied, (iv) Molar heat capacity of the gas R = 8.31 J mol–1 K–1

SolutionP1 = 4 × 105 Nm–2 P2 = 8 × 10+5 Nm–2, V1 = 0.2 m3, V2 = 0.5 m3

(i) Work done by the gas = Area under P–V graph (Area ABCDEA)

(10

× N

/m)

52

4

0

8 B

C

0.50.2

A

V(m )3

P

DE

12

= (AE +BD) × AC 12

= (4 × 105 + 8 × 105) × (0.5–0.2)

12

= × 12 × 105 × 0.3 = 1.8 × 105 J

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

51E

Pre-Medical : PhysicsALLEN(ii) Increase in internal energy DU = CV (T2 – T1)

VCR

= R(T2–T1) VC

R= (P2V2 – P1V1)

For monoatomic gas CV 3

R2

=

\ DU 3

2= [(8 × 105 × 0.5) – (4 × 105 × 0.2)]

3

2= [4 × 105 – 0.8 × 105 ] = 4.8 × 105 J

(iii) Q = DU + W = 4.8 × 105 + 1.8 × 105 = 6.6 × 105 J

(iv) C Q QR

T R T= =

mD m D=

-2 2 1 1

QR(P V P V )

5

5

6.6 10 8.311 3.2 10

´ ´=

´ ´= 17.14 J/mol K

Illustration 54.

As shown in figure when a system is taken from state a to state b, along the path

P

V

c b

da

a ® c ® b, 60 J of heat flow into the system, and 30 J of work is done :

(i) How much heat flows into the system along the path a ® d ® b if the work is 10 J.

(ii) When the system is returned from b to a along the curved path, the work done by the system is

–20 J. Does the system absorb or liberate heat, and how much ?

(iii) If, Ua = 0 and Ud = 22 J, find the heat absorbed in the process a ® d and d ® b.

Solution

For the path acb DU = Q – W = 60 – 30 = 30 J or Ub – Ua = 30 J

(i) Along the path adb Q = DU + W = 30 + 10 = 40 J

(ii) Along the curved path ba Q = (Ua – Ub) + W = (–30) + (–20) = –50 J, heat liberates from system

(iii) Qad = Ud – Ua + Wad

but Wad = Wadb – Wdb = 10 – 0 = 10 Hence Qad = 22 – 0 + 10 = 32 J

and Qdb = Ub – Ud + Wdb = 30 – 22 + 0 = 8 J

Illustration 55.

1 kg of water at 373 K is converted into steam at same temperature. Volume of 1 cm3 of water becomes

1671 cm3 on boiling. What is the change in the internal energy of the system if the latent heat of vaporisation

of water is 5.4 × 105 cal kg–1?

Solution

Volume of 1 kg of water = 1000 cm³ = 10–3 m³, Volume of 1 kg of steam = 10³ × 1671 cm³ = 1.671 m³

Change in volume, DV = (1.671 – 10–3) m³ = 1.670 m³ , Pressure, P = 1 atm. = 1.01 × 105 N m–2

In expansion work done, W = PDV = 1.01 × 105 × 1.67 J =51.686 15

4.2´

cal = 4.015 × 104 cal

But DU = Q – W (first law of thermodynamics) or DU = (5.4 × 105 – 0.4015 × 105) cal = 4.9985 × 105 cal

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

52 E

Pre-Medical : Physics ALLENBEGINNER'S BOX-7

1. One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in fig. calculate.

(a) The work done by the gas.

(b) The heat rejected by the gas in the path CA and heat absorbed in the path AB.

(c) The net heat absorbed by the gas in the path BC.

2. Calculate the heat supplied in figure shown for complete cycle.

3. Given figure shows a P–V graph of the thermodynamic behaviour of an ideal gas. Find out work in the processes

A® B, B ® C, C® D and D ® A from this graph.

4. At 8 atm pressure a monoatomic gas expands from 2 litre to 5 litre then calculate the following

(a) Work done by the gas

(b) Increase in internal energy

(c) Amount of heat supplied

5. A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process.

Calculate the value of J.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

53E

Pre-Medical : PhysicsALLEN6. DIFFERENT PROCESSES

6.1 Isometric or Isochoric Process

Isochoric process is a thermodynamic process that takes place at constant volume of the system, but pressure

and temperature varies for change in state of the system.

Equation of state P/T = constant (P and T are variable, V is constant)

Work done In this process volume remains constant so dV = 0 Þf

i

V

VW PdV 0= =ò

Form of first Law Q = DU

It means whole of the heat supplied is utilized for change in internal

energy of the system. Q = DU = µ Cv DT

Slope of the P–V curve dPdV

= ¥V

P

dPdV

= ¥

V

P

dVdP

= 0

Specific heat or molar heat capacity at constant volume (CV)

The quantity of heat required to raise the temperature of 1 gram mole of gas through 1 °C at constant volume

is equal to the specific heat at constant volume.

l A gas enclosed in a cylinder having rigid walls and a fixed piston. When heat is added to the gas, there would

be no change in the volume of the gas.

l When a substance melts, the change in volume is negligibly small. So, this may be regarded as a nearly

isochoric process.

l Heating process in pressure cooker is an example of isometric process.

6.2 Isobaric Process

Isobaric process is a thermodynamic process that takes place at constant pressure, but volume and temperature

varies for change in state of the system.

l Equation of state V/T = Constant or V µ T

l Work done In this process pressure remains constant \ dP = 0

Work done f

i

V

f iVW PdV P(V V )= = -ò

Þ Fraction of Heat given which –

(i) is converted into internal energy U 1Q

D=

D g(ii) does work against external pressure

W 11

QD

= -D g

l Form of first Law Q = DU + P (Vf – Vi)

µ Cp (Tf – Ti) = µ Cv (Tf – Ti) + P(Vf – Vi)

It is clear that heat supplied to the system is utilized for :

(i) Increasing internal energy and (ii) Work done against the surrounding atmosphere.

• Slope of the PV curve : isobaric

dP0

dVæ ö =ç ÷è ø

V

P V

P

dVdP

= ¥dPdV

= 0

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

54 E

Pre-Medical : Physics ALLEN• Specific heat at constant pressure (CP)

The quantity of heat required to raise the temperature of 1 gram mole gas through 1°C at constant pressure

is equal to the specific heat.

Ex. Heating of water at atmospheric pressure. • Melting of solids and boiling of liquids at atmospheric pressure.

6.3 Isothermal ProcessIn this process pressure and volume of system change but temperature remains constant. In an isothermal

process, the exchange of heat between the system and the surroundings is allowed. Isothermal process is

carried out by either supplying heat to the substance or by extracting heat from it. A process has to be extremely

slow to be isothermal.

• Equation of state : P V = constant (µRT) (T is constant)

• Work DoneConsider µ moles of an ideal gas, enclosed in a cylinder, at absolute temperature T, fitted with a frictionless

piston. Suppose that gas undergoes an isothermal expansion from the initial state (P1, V1) to the final state

(P2, V2). Work done : 2

1

V

VW PdV= ò Q PV = µRT then

RTP

Vm

=

\ 2

1

V

V

µRTW dV

V= ò =

2

1

V

V

dVµRT

Vò = [ ] 2

1

V

e VµRT log V = µRT [loge V2 – loge V1] = 2

e1

VµRT log

Vé ùê úë û

Þ W = 2.303µRT log101

2

PP

é ùê úë û

[Q P1V1 = P2V2]

Form of First LawThere is no change in temperature and internal energy of the system depends on temperature only

So DU = 0, Q = 2.303 µRT log102

1

VV

é ùê úë û

It is clear that Whole of the heat energy supplied to the system is utilized by the system in doing external

work. There is no change in the internal energy of the system.

Slope of the isothermal curve

For an isothermal process, PV = constant

Differentiating, PdV + VdP = 0 Þ VdP = – PdV Þ dP PdV V

= -

P

V

isotherm

hyperbola

PV = constantPi

Pf

VfVi

Slope of isothermal curve, isothermal

dP PdV V

é ù = -ê úë û

For a given system :

• The product of the pressure and volume of a given mass of a perfect gas remains constant in an isothermal process.

• Boyle's law is obeyed in an isothermal process.

• A graph between pressure and volume of a given mass of a gas at constant temperature is known as isotherm

or isothermal curve of the gas.

• Two isotherms for a given gas at two different temperatures T1 and T2 are shown in figure.

P

V

T1

T2

T > T2 1

so T > T2 1at const P; V T µ

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

55E

Pre-Medical : PhysicsALLEN• The curves drawn for the same gas at different temperatures are mutually parallel and do not cut each other.

• If two isotherms intersect each other at a single point we get same value of P and V at intersection point.

PV = µ RT1 for temperature T1 and PV = µ RT2 for temperature T2.

It means T1 = T2 which is not possible.

• An ideal gas enclosed in a conducting cylinder fitted with a conducting piston.

• Let the gas be allowed to expand very–very slowly. This shall cause a very slow cooling of the gas, but

heat will be conducted into the cylinder from the surrounding. Hence the temperature of the gas remains

constant. If the gas is compressed very–very slowly, heat will be produced, but this heat will be conducted

to the surroundings and the temperature of the gas shall remain constant.

• The temperature of a substance remains constant during melting. So, the melting process is an isothermal process.

• Boiling is an isothermal process, when a liquid boils, its temperature remains constant.

• If sudden changes are executed in a vessel of infinite conductivity then they will be isothermal.

6.4 Adiabatic Process

It is that thermodynamic process in which pressure, volume and temperature of the system do change but

there is no exchange of heat between the system and the surroundings.

A sudden and quick process will be adiabatic since there is no sufficient time available for exchange of heat

so process adiabatic.

Equation of state : PV = µRT

Equation for adiabatic process PVg = constant, TV

g–1 = constant, Tg p1–g = constant

Work done

Let initial state of system is (P1, V1, T1) and after adiabatic change final state of system is (P2, V2, T2) then

we can write P1 V1g = P2 V2

g = K (here K is const.)

So2

1

V

VW PdV= ò 2

1

V

VK V dV-g= ò

2

1

V1

V

VK

1

g- +æ ö= ç ÷g- +è ø

1 12 1

KV V

( 1)g g- + - +é ù= -ë ûg- + ( )1 1 2 2K P V P Vg g= =Q

Þ W 1 1 1 1 2 2 2 2

1P V V .V P V V .V

( 1)g gg - g -é ù= -ë ûg -

=1

( 1)g - [P1V1– P2V2] Þ ( )1 2

µRW T T

( 1)= -

g - (Q PV = µRT)T)

Form of first law : dU = – dW

It means the work done by an ideal gas during adiabatic expansion (or compression) is on the cost of change

in internal energy proportional to the fall (or rise) in the temperature of the gas.

If the gas expands adiabatically, work is done by the gas. So, Wadia is positive.

The gas cools during adiabatic expansion and T1 > T2.

If the gas is compressed adiabatically, work is done on the gas. So, Wadia is negative.

The gas heats up during adiabatic compression and T1 < T2.

Slope of the adiabatic curve

For an adiabatic process, PVg = constant

V

P

(Adiabatic expansion of gas)

polyatomicdiatomic

monoatomic

g= 1.33

g= 1.4g= 1.67

Differentiating, g P Vg – 1 dV + Vg dP = 0

Þ Vg dP = – g PVg – 1 dV Þ 1dP PV P P

dV V VV

g -

g

g æ ö= - = -g = g -ç ÷è ø

Slope of adiabatic curve, adiabatic

dP PdV V

gé ù = -ê úë û

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

56 E

Pre-Medical : Physics ALLENMagnitude of slope of adiabatic curve is greater than the slope of isotherm

= g = gadia iso

dP P dPdV V dV Þ

slope of adiabatic changes=

slope of isothermal changesg

V

Pisothermal

(expansion of gas)

adiabatic

Since g is always greater than one so an adiabatic curve is steeper than an

isothermal curve

Examples of adiabatic process

• A gas enclosed in a thermally insulated cylinder fitted with a non–conducting piston. If the gas is compressed

suddenly by moving the piston downwards, some heat is produced. This heat cannot escape from the cylinder.

Consequently, there will be an increase in the temperature of the gas.

• If a gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature

of the gas.

• Bursting of a cycle tube.

• Propagation of sound waves in a gaseous medium.

• In diesel engines burning of diesel without spark plug is done due to adiabatic compression of diesel vapour

and air mixture

Free Expansion

Take a thermally insulated bottle with ideal gas at some temperature T1 and, by means of a pipe with a stopcock,

connect this to another insulated bottle which is evacuated. If we suddenly open the stopcock, the gas will

rush from the first bottle into the second until the pressures are equalized.

free expansion of a gas

Vaccum (P = 0)

Experimentally, we find that this process of free expansion does not change the temperature of the gas –

when the gas attains equilibrium and stops flowing, the final temperature of both botles are equal to the

initial temperature T1. This process is called a free expansion.

The change in the internal energy of the gas can be calculated by applying the first law of thermodynamics

to the free–expansion process.

The process is adiabatic because of the insulation, So Q = 0.

No part of the surroundings moves so the system does no work on its surroundings.

• For ideal gas (dW)ext. = Work done against external atmosphere = P dV = 0

(Q P = 0) Þ because vacuum

(dW)int.= Work done against internal molecular forces = 0

dQ = dU + dW Þ 0 = dU + 0

The internal energy does not change dU = 0 So U and T are constant.

The initial and final states of this gas have the same internal energy.

Which implies that the internal energy of an ideal gas does not depend on the volume at all.

The free–expansion process has led us to the following conclusion :

The internal energy U(T) of an ideal gas depends only on the temperature.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

57E

Pre-Medical : PhysicsALLENRelation between degrees of freedom and specific heat of gas

Energy related with each degree of freedom for a gas molecule =1/2 kT, Energy related with all degree of

freedom for a gas molecule = f/2 kT

Internal energy of one mole of ideal gas (total K.E.) U = f/2 kT for Isometric process (volume constant) dW=0

By first law of thermodynamics dQ = dW + dU Þ CV dT = dU Þ CV = dUdT

= = =g -V

dU f RC R

dT 2 1, P V

f RC C R 1 R

2 1gé ù= + = + =ê ú g -ë û

and P

V

C 21

C fg = = +

=g -V

RC

1,

P

RC

1g

=g -

and g = +2

1f

General expression for C (CP or CV) in the process PVx = constant = +g - -

R RC

1 1 x

For isobaric process P = constant so x = 0 \ = = + = +g -P V

RC C R C R

1

For isothermal process, PV = constant so x = 1 \ C = ¥

For adiabatic process PV g = constant so x = g \ C = 0

Values of f, U, CV, CP and g for different gases are shown in table below.

Atomicity of gas f Cv CP g

Monoatomic 3 3

R2

5R

2

51.67

3=

Diatomic, Triatomic and Triatomic linear (at normal temperature) 5

5R

2

7R

2

71.4

5=

Poly atomic Triangular Non-linear 6 6

R 3R2

=

8R 4R

2=

41.33

3=

• 1 < g < 2

• If atomicity of gases is same U, CP, CV and g is same for gas mixture.

• If in a gas mixture gases are of different atomicity, then for gas mixture g changes according to following

condition. Diatomic g1 < mixture < g2 mono atomic where g1 < g2

• If 'f' is the degree of freedom per molecule for a gas, then

* Total energy of each molecule =fkT2

* Total energy per mole of gas = 0

f fN kT RT

2 2=

• According to kinetic theory of gases, the molecule are not interacting with each other. So potential energy

is zero and internal energy of gas molecules is only their kinetic energy.

• For 'm' mole of a gas : Internal energy at temperature T is V

f R TU C T

2m

= = m

• Change in internal energy is given by V

fRdU (dT) C dT

2m

= = m . This change is process independent.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

58 E

Pre-Medical : Physics ALLENCP is greater than CV

If a gas is heated at constant volume, the gas does no work against external pressure. In this case, the whole

of the heat energy supplied to the gas is spend in raising the temperature of the gas.

If a gas is heated at constant pressure, its volume increases. In this case, heat energy is required for the following

two purpose :

(i) to increase the volume of the gas against external pressure.

(ii) To increase the temperature of 1 mole of gas through 1 K.

Thus, more heat energy is required to raise the temperature of 1 mole of gas through 1 K when it is heated

at constant pressure than when it is heated at constant volume. \ CP > CV

The difference between CP and CV is equal to thermal equivalent of the work done by the gas in expanding

against external pressure.

Mayer's formula : CP – CV = R

Q At constant pressure dQ = µCPdT, dU = µCVdT & dW = PdV = µRdT

Now from first law of thermodynamics dQ = dW + dU

Þ µCPdT = µRdT + µCVdT Þ CP = R + CV Þ CP – CV = R

GOLDEN KEY POINTS

l When a gas expands its volume increases, then final pressure is less for adiabatic expansion. But, when a

gas compresses its volume decreases, then the final pressure is more in case of adiabatic compression.

V

P

isobaric

isothermal

adiabatic

isobaric

adiabatic

isothermal

V

PW < W < WIB IT AD

P P > P > Pfinal AD IT IB

DP P P P AD IT IB > > D D DAD

IT

IB

V/2 VV

PW > W > WIB IT AD

P P > P > Pfinal IB IT AD

DP P P P AD IT IB > > D D DIB

IT

AD

2VV

First Law of Thermodynamics Applied to Different Processes

Process Q DU W

Cyclic W 0 Area of the closed curve

Isochoric DU µCvDT (µ mole of gas) 0

Isothermal W 0 f

ei

VRTlog

Vé ù

m ê úë û

= ie

f

PRT log

Pé ù

m ê úë û

Adiabatic 0 –W f iµR(T T )

1

-

- g

Isobaric µCPDT mCVDT P (Vf – V

i) = µR(T

f – T

i)

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

59E

Pre-Medical : PhysicsALLENIllustrations

Illustration 56.

An ideal gas has a specific heat or molar heat capacity at constant pressure CP =5R/2. The gas is kept in

a closed vessel of volume 0.0083 m3 at a temperature of 300K and a pressure of 1.6 × 106 Nm–2. An

amount of 2.49 × 104 J of heat energy is supplied to the gas. Calculate the final temperature and pressure

of the gas.

Solution

CV = CP – R = 5R 3R

R2 2

- = , DV = 0, T1 = 300 K, V = 0.0083 m3, P1 = 1.6 × 106 Nm–2

From first law of thermodynamics Q = DU + PDV Þ DU = Q = 2.49 × 104 J

From gas equationPV

nRT

=61.6 10 0.0083 16

8.3 300 3´ ´

= =´

\ DU = nCVDT Þ 4

v

U 2.49 10 6T

C 3 8.3 16D ´ ´

D = =m ´ ´ = 375 K

Final temperature = 300 + 375 = 675K

According to pressure law P µ T Þ2 2

1 1

P TP T

= Þ 2

2 11

TP P

T= ´

61.6 10 675300

´ ´= = 3.6 × 106 Nm–2

Illustration 57.

5 moles of oxygen is heated at constant volume from 10°C to 20°C. What will be change in the internal

energy of the gas? The gram molecular specific heat of oxygen at constant pressure.

CP = 8 cal/mole and R = 8.36 joule/mole °C.

Solution

Q CV = CP – R = 8 – 2 = 6 cal/mole °C

\ Heat absorbed by 5 moles of oxygen at constant volume

Q = mCV D T = 5 × 6 (20 – 10) = 30 × 10 = 300 cal

At constant volume D V = 0. \ DW = 0

\ From first law of thermodynamics Q = DU + W Þ 300 = DU + 0 Þ DU = 300 cal

Illustration 58.

What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise the temperature

by 45°C at constant pressure. Molecular mass of N2 = 28, R = 8.3 J mol–1 K–1.

Solution

Here m = 2 × 10–2 kg, Þ m = 2

3W

M 2 10 5M 728 10

-

-´

= =´

& P

7C R

2= \ Q = mCP DT =

5 78.3 45 933.75

7 2´ ´ ´ = J

Illustration 59.

Two moles of a gas at 127°C expand isothermally until its volume is doubled. Calculate the amount of work

done.

Solution

m = 2, T = 127 + 273 = 400K, 2

1

VV

=2

From formula W = 2.3026 mRT log10 2

1

VV

= 2.3026 × 2 × 8.3 × 400 × log10 2

= 2.3026 × 2 × 8.3 × 400 × 0.3010 » 4.6 × 103J

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

60 E


Figure shows a process ABCA performed on an ideal gas. Find the net heat given to the system during the

process.V

T1 T2 T

V2

V1A B

C

Solution

Since the process is cyclic, hence the change in internal energy is zero.

The heat given to the system is then equal to the work done by it.

The work done in part AB is W1 = 0 (the volume remains constant). The part BC represents an isothermal

process so that the work done by the gas during this part is W2 = mRT2 ln 2

1

VV

During the part CA : V µ T So, V/T is constant and hence, RT

PV

m= is constant

The work done by the gas during the part CA is W3 = P(V1 – V2) = mRT1 – mRT2 = – mR (T2 – T1).

The net work done by the gas in the process ABCA is W = W1 + W2 + W3 = 22 2 1

1

VR T n (T T )

V

é ùm - -ê ú

ë ûl

The same amount of heat is given to the gas.

Illustration 61.

During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially

at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2V.

Solution

By gas equation PV = mRT and VP2 = constant on eliminating P

é ù= mê ú

ë û

AV RT

VÞ

m=

RV T

A\

é ù= ê ú

ë û1 1

22

V TTV

Þ =V T

T '2VÞ T' = ( )2 T

Illustration 62.

2m3 volume of a gas at a pressure of 4 × 105 Nm–2 is compressed adiabatically so that its volume becomes

0.5m3. Find the new pressure. Compare this with the pressure that would result if the compression was isothermal.

Calculate work done in each process. g = 1.4

Solution

V1 = 2m3, P1 = 4 × 105 Nm–2, V2 = 0.5m3

In adiabatic process 1 1P Vg = 2 2P Vg Þ P2 = 4 × 1051.4

20.5

é ùê úë û

= 4 × 105 (4)1.4 = 2.8 × 106 Nm–2

In isothermal process P1V1 = P2V2 Þ 5

1 12

2

P V 4 10 2P

V 0.5´ ´

= = = 1.6 × 106 Nm–2.

Now work done in adiabatic process 1 1 2 2P V P V

W1

-=

g - 5 6

6(4 10 2 2.8 10 0.5)1.48 10 J

1.4 1´ ´ - ´ ´

= = - ´-

Work done in isothermal process W = 2.303mRT log 2

1

V

V = 2.3026P1V1 log 2

1

V

V

= 2.3026 × 4 × 105 × 2 × log0.52.0

é ùê úë û

= 2.3026 × 4 × 105 × 2log14

æ öç ÷è ø = –1.1 × 106 J

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

61E


Two samples of a gas initially at same temperature and pressure are compressed from a volume V to V2

.

One sample is compressed isothermally and the other adiabatically. In which sample is the pressure greater?

Solution

Let initial volume, V1 = V and pressure, P1 = P , final volume, V2 = V2

and final pressure, P2 = ?

For isothermal compression P2V2 = P1V1 or 1 12

2

P V PVP 2P

VV2

= = =

For adiabatic compression P2' = P1 1

2

VV

gé ùê úë û

Þ 2

VP ' P 2 P

V /2

ggé ù= =ê úë û

Þ P2´ = 2gP g > 1 \ 2

g > 2 and P2' > P2

Pressure during adiabatic compression is greater than the pressure during isothermal compression.

Illustration 64.

Plot P – V , V – T graph corresponding to the P–T graph for an ideal

gas shown in figure. Explain your answers.

Solution

(P - V curve)

D

A

CB

V

T(V - T curve)

For process AB ; T = constant so P µ 1V

For process BC P = constant so V µ T

For process CD ; T = constant so V µ 1P

For process DA P = constant so V µ T

Illustration 65.

The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate its atomic weight,

[R = 2cal/mol K]

Solution

As argon is monoatomic, its molar specific heat at constant volume will be

V

3 3C R 2 3

2 2= = ´ = cal/mol K, CV =Mw cV and cV = 0.075 cal/g K

So 3 = Mw × 0.075 Þ Mw = 3

0.075 = 40 gram/mole

Illustration 66.

Two moles of helium gas (g = 5/3), assumed ideal, are initially at 27°C and occupy a volume of 20 litres.

The gas is first expanded at constant pressure till its volume is doubled. It then undergoes an adiabatic change

until the temperature returns to its initial value. Determine the final pressure and volume of the gas. Also,

calculate the work done under isobaric and adiabatic processes. [R = 8.3 J mol–1 K–1] [AIPMT (Mains) 2005]

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

62 E

Pre-Medical : Physics ALLENSolution

For process AB (P = constant) T2 = 2T1 = 600K(QV2= 2V1)

For process BC TVg–1 = constant Therefore (2T1) (V2)5/3–1

= (T1) (V3)5/3–1

Þ V3 = (V2) (2)3/2 = 2 2 V2 = 80 2 × 10–3 m3 = 80 2 litres

P3 = mRT

V3

3 =

( ) ( . ) ( )2 83 300

80 2 10 3´ - = 0.44 × 105 N/m2

Work done under isobaric process = PDV = P1(V2 – V1)

= mRT

VV V1

12 1 3

3 32 83 300

20 1040 10 20 10( )

.( )- =

´ ´´

´ - ´-- - = 4980J

Work done under adiabatic process = m

gR T T( )

( )2 3

1-

- =

2 8 3 3005 3 1

´ ´-

. ( )

b g = 7470J

BEGINNER'S BOX-8

1. 32 gram of oxygen gas at temperature 27°C is compressed adiabatically to 1/3 of its initial volume. Calculate

the change in internal energy ? (g = 1.5 for oxygen)

2. How much heat energy should be added to the gaseous mixture consisting of one gram of hydrogen and

one gram helium to raise its temperature from 0°C to 50°C.

(a) at constant volume and

(b) at constant pressure, assuming both the gases to be ideal.

R = 2 cal/mol-K, gH2

= 1.41, gHe = 1.67

3. Match the column–I with column–IIColumn – I Column – II

(A) (p) Isometric process

(B) (q) Adiabatic process

(C) (r) Isothermal process

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

63E

Pre-Medical : PhysicsALLEN4. Match the column–I with column–II

Column–I Column–II

(A) Isobaric process (p) No heat exchange

(B) Isothermal process (q) Constant pressure

(C) Isoentropic process (r) Constant internal energy

(D) Isochoric process (s) Work done is zero

5. Match the column–I with column–II

Column–I Column–II

(A) Adiabatic expansion (p) Q = 0

(B) Adiabatic free expansion (q) DU = 0

(C) Isobaric expansion (r) W = 0

(D) Isothermal expansion (s) Q = W + DU

(E) Isochoric cooling

6. An ideal gas expands isothermally along AB and does 700 J of work.

(a) How much heat does the gas exchange along AB.

(b) The gas expands adiabatically along BC and does 400 J of work. When

the gas returns to A along CA, it exhusts 100 J of heat to its sur-

roundings. How much work is done on the gas along this path.

7. For one mole of an ideal gas at 300 K occupies a volume of 0.36 m3 at

a pressure of 2 atm. The gas expands adiabatically until its volume becomes

1.44 m3. Next gas is compressed isobarically to its original volume. Finally,

the pressure is increased isochorically until the gas returns to its

initial state. (see figure).

(a) Determine the temp. at the point B and C

(b) The work done during the process A to B

(Assume g = 1.5, R = 8 J

mol K-)

8. Following figure is given for 1 mole gas.

Then find out net work done in the cyclic process.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

64 E

Pre-Medical : Physics ALLENEFFECIENCY OF A CYCLE (h) :

h =+

total Mechanical work done by the gas in the whole processHeat absorbed by the gas (only ve)

= area of the P-V cycle

Heat injected into the system

HEAT ENGINE

Heat engine is a device which converts heat into work.

Q1

cold reservoir T2

Q2

Wworking substance

sink

hot reservoir T1

source

Three parts of a heat engine:

(i) Source of high temperature reservoir at temperature T1

(ii) Sink of low temperature reservoir at temperature T2

(iii) Working substance.

In a cycle of heat engine the working substance extracts heat Q1 from source, does

some work W and rejects remaining heat Q2 to the sink.

Efficiency of heat engine 1

work done (W)heat taken fromsource (Q )

h = Þ-1 2

1

T TT

= 1 2

1

Q QQ-

REFRIGERATOR

It is inverse of heat engine. It extracts heat (Q2) from a cold reservoir,

some external work W is done on it and rejects heat (Q1) to hot reservoir.

The coefficient of performance of a refrigerator (C.O.P. or b).

eat extracted from cold reservoirH

ork done on refrigeratorW b =

2 2

11 2

2

Q Q 1QW Q Q 1Q

= = =- -

Q = Q + W1 2

Hot reservoir T1

Cold reservoir T2

Q2

W

For Carnot reversible refrigerator 1 1

2 2

Q TQ T

= \ 2

1 1

2 2

Q 1 1W Q T

1 1Q T

b = = =é ù é ù

- -ê ú ê úë û ë û

Þ b = 2

1 2

T

T T-

Relation between b and h : Q1

WQ

h = [Efficiency of heat engine]

(Refrigerator and heat engine 2QW

b = [C.O.P. of refrigerator]

are maintained at same thermal reservoirs)and Q1 = Q2 + W

then 1

1b = -h

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

65E

Pre-Medical : PhysicsALLENSECOND LAW OF THERMODYNAMICS

The first law of thermodynamics is the principle of conservation of energy. Common experience shows that

there are many conceivable processes that are perfectly allowed by the First law and yet are never observed.

For example, nobody has ever seen a book lying on a table jumping to a height by itself. Would be possible

if the principle of conservation of energy were the only restriction. This principle, which disallows many

phenomena consistent with the first law of thermodynamics is known as the second law of thermodynamics.

The Second Law of Thermodynamics gives a fundamental limitation to the efficiency of a heat engine and

the co-efficient of performance of refrigerator.

Second Law of Thermodynamics

Kelvin Planck Statement

No process is possible whose sole result is the absorption of heat from a reversoir and the complete conversion

of the heat into work.

Claussius Statement

No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.

It can be proved that the two statements above are completely equivalent.

The second law of thermodynamics is a general principle which places constrains upon the direction of heat

transfer and the attainable efficiencies of heat engines.

Entropy Statement of the second Law of Thermodynamics

The entropy of the Universe never decrease. It either increases (for irreversible processes) or remains the

same (for reversible processes)

REVERSIBLE AND IRREVERSIBLE PROCESS

(i) Reversible Process

A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings.

Both the system and the surroundings are returned to their initial states at the end of the reverse process.

Conditions for a Reversible process

(a) The process should take place very slowly, i.e. quasi-statically, i.e. seemingly static (from the Latin word

qasi meaning ‘as if’) so that it satisfies the following requirements at each stage of the process.

• The system should be in mechanical equilibrium.

• The system should be in thermal equilibrium.

• The system should be in chemical equilibrium, i.e., no new products should be formed.

(b) There should be no friction losses etc. It should be remembered that a complete reversible process

or cycle of operations is only an ideal case. In actual practice, there is always a loss of heat due to friction,

conduction and radiation. However, it is possible to approximate reversible processes through carefully controlled

procedures, but they can never be achieved.

Examples

• Frictionless pendulum.

• Quasi-equilibrium expansion and compression of a gas.

(b) Irreversible Process

A process that is not reversible is called an irreversible process.

The spontaneous processes occuring in nature are irreversible. In fact irreversibility is a rule rather than an

exception in nature.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

66 E

Pre-Medical : Physics ALLENExamples

(i) Sudden unbalanced expansion of a gas.

(ii) Heat produced by friction.

(iii) Heat generated when a current flows through an electric resistance in any direction.

(iv) Cooking gas leaking from a cylinder.

(v) Diffusion of liquids or gases.

(vi) Breaking of an egg.

(vii) The growth of a plant.

A system can be restored to its initial state following a process, regardless of whether the process is reversible

or irreversible. But for reversible process, this restoration is made without leaving any net change on the

surroundings whereas for irreversible processes, the surroundings usually do some work on the system and

therefore will not return to their original state.

ENTROPY AND THE SECOND LAW OF THERMODYNAMICS

To have a better understanding of thermodynamic process and the second and third laws of thermodynamics,

let us introduce the concept of entropy. It was introduced by clausius in 1865.

A quantity that denotes the amount of disorder is called entropy and is denoted by S.

The total energy always conserved in any process but the total entropy always increases or remains the same

in any process (ie., the disorder increases or remain constant)

dS = dQT

Þ DS = dQT

óôõ .....(1)

For reversible processes, DS = 0. For irreversible process DS > 0

A process where DS < 0 is not possible.

3. Carnot cycle

Carnot devised an ideal engine which is based on a

reversible cycle of four operations in succession :

(i) Isothermal expansion, A ® B

V

P1A

B

C

D

T2

T1

P2

P4

P3

P

V3V2V4V1

Q1

Q2

on source

on stand

on stand

on sink(ii) Adiabatic expansion, B ® C

(iii) Isothermal compression C ® D

(iv) Adiabatic compression. D ® A

Main parts of Carnot's engine are

l Source of heat :

It is a hot body of very large heat capacity kept at a constant high temperature T1. Its upper surface is perfectly

conducting so that working substance can take heat from it.

l Mechanical arrangements and working substance :

It is a hollow cylinder whose walls are perfectly non–conducting and its base is perfectly conducting fitted with

non–conducting piston. This piston move without any friction. Ideal gas enclosed in cylinder as a working substance.

l Heat sink :

It is a cold body at low temperature T2. It is a body of large heat capacity and its upper surface is highly

conducting so that working substance can reject heat to it.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

67E

Pre-Medical : PhysicsALLENl Stand :

It is made by perfectly insulating material. When cylinder placed on it working substance can expended or

compressed adiabatic.

l Working :

A set of reversible processes through them working substance is taken back to initial condition to get maximum

work from this type of ideal engine.

Processes of Carnot's cycle can be denoted by an indicator diagram.

l Isothermal expansion A ® B

Initially the cylinder is taken to be in thermal equilibrium with the high temperature

T1, this is initial state of working substance denoted by point A (P1, V1, T1).

After that the piston is allowed to move outward slowly. With the movement

of the piston. The process is very slow so that it is isothermal. Heat from reservoir

flows through the base of cylinder into the gas so temperature of the gas remains

T1. Gas expand and receive heat Q1 from source and gets state B(P2, V2, T1)

This heat input Q1 to the gas from path A to B is utilized for doing work W1.

By path A to B the heat input to the gas = the work done against the external pressure.

2 2

1 1

V V1 2

1 1 1V V1

µRT VW Q PdV dV µRT n

V V= = = =ò ò l

l Adiabatic expansion B ® C

Now cylinder is put in contact with a non–conducting stand and piston is allowed to

Q=0

move outward, because no heat can enter in or leave out so the expansion of gas

is adiabatic. The temperature falls to T2K and gas describes the adiabatic from B

to point C (P3, V3, T2). During this expansion more work is done (W2) at the expense

of the internal energy.

Work done in adiabatic path BC is ( )2 1 2

µRW T T

1= -

g -l Isothermal compression C ® D

Now the gas cylinder is placed in contact with sink at temperature T2. The piston is

Q2

sink

moved slowly inward so that heat produced during compression passes to the sink.

The gas is isothermally compressed from C to point D (P4, V4, T2). The heat rejected

Q2 to the cold reservoir (sink) at T2 occurs over this path. Amount of work done on

gas W3 = amount of heat rejected to the sink

Q2 = W3 = mRT2ln 4

3

VV

æ öç ÷è ø

Þ Q2 = µRTT2ln4

3

VV

æ öç ÷è ø

l Adiabatic compression D ® A

Q=0

The cylinder is removed from the sink and is put in contact with insulating

stand now piston moves inward. Heat is not allowed to go out and it increases

the internal energy of the system. Now work is done on the gas during adiabatic

compression from state D to initial point A (P1, V1, T1). No heat exchanges

occur over the adiabatic path. Work done on the system W4 = 2 1

µR(T T )

1-g -

This cycle of operations is called a Carnot cycle.

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

68 E

Pre-Medical : Physics ALLENIn first two steps work is done by engine W1 and W2 are positive

In last two steps work is done on gas W3 and W4 are negative

The work done in complete cycle W = the area of the closed part of the P–V cycle.

W = W1 + W2 + W3 + W4

\ = + - + - -g g- -l l

2 41 1 2 2 2 1

1 3

V VµR µRW µRT n (T T ) µRT n (T T )

V 1 V 1 2 4

1 21 3

V VµRT n µRT n

V V= +l l

Efficiency of Carnot Engine, 1

W

Qh = Þ

2 41 2

1 3

21

1

V VµRT n µRT n

V VV

µRT nV

+

h =

l l

l

B to C and D to A are adiabatic paths

so T1V2(g – 1) = T2 V3

(g – 1) and T1V1(g – 1) = T2 V4

(g – 1) Þ2 3

1 4

V VV V

=

Þ 1 2

1

T TT-

h = = 1 2

1

Q QQ-

Þ 2 2

1 1

Q T1 1

Q Th = - = - 1 2

1 2

Q QT T

=

1 2

1

T T100%

T-

h = ´ Þ1 2

1

Q Q100%

Q-

h = ´

The efficiency for the Carnot engine is the best that can be obtained for any heat engine. The efficiency

of a Carnot engine is never 100% because it is 100% only if temperature of sink T2 = 0 which is impossible.

Carnot Theorem

No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same

hot and cold reservoirs.2 2

R I1 1

T Q1 1

T Qh > h Þ - > -


A carnot engine working between 400 K and 800 K has a work output of 1200 J per cycle. What is the

amount of heat energy supplied to the engine from source per cycle?

Solution

W = 1200 J, T1 = 800 K, T2 = 400 K \ h = 1 – 2

1 1

T WT Q

= Þ 1

400 12001

800 Q- =

Þ 0.5 = 1

1200Q

Heat energy supplied by source 1

1200Q

0.5= = 2400 joule per cycle

Illustration 68.

The temperatures T1 and T2 of the two heat reservoirs in an ideal carnot engine are 1500°C and 500°C

respectively. Which of the following : increasing T1 by 100°C or decreasing T2 by 100°C would result in

a greater improvement in the efficiency of the engine?

Solution

T1 = 1500°C = 1500 + 273 = 1773 K and T2 = 500°C = 500 + 273 = 773 K.

The efficiency of a carnot's engine 2

1

T1

Th = -

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

69E

Pre-Medical : PhysicsALLENWhen the temperature of the source is increased by 100°C, keeping T2 unchanged, the new temperature

of the source is T´1 = 1500 + 100 = 1600°C = 1873 K.

The efficiency becomes 2

1

T 773´ 1 1 0.59

T ' 1873h = - = - =

On the other hand, if the temperature of the sink is decreased by 100°C, keeping T1 unchanged, the new

temperature of the sink is T´2 = 500 – 100 = 400°C = 673 K. The efficiency now becomes

2

1

T´ 673´´ 1 1 0.62

T 1773h = - = - =

Since h´´ is greater than h´, decreasing the temperature of the sink by 100°C results in a greater efficiency

than increasing the temperature of the source by 100°C.

Illustration 69.

A heat engine operates between a cold reservoir at temperature T2 = 300 K and a hot reservoir at temperature

T1. It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle.

What could be the minimum temperature of hot reservoir?

Solution

Work done by the engine in a cycle is W = 200 – 120 = 80 J. h = W 80

0.4Q 200

= =

From carnot's Theorem 2

1 1

T 3000.4 1 1

T T£ - = - Þ

1

3000.6

T£ Þ 1

300T

0.6³ Þ T1 ³ 500

Illustration 70.

A carnot engine works as a refrigrator between 250 K and 300 K. If it receives 750 cal of heat from the

reservoir at the lower temperature. Calculate the amount of heat rejected at the higher temperature.

Solution

T1 = 300 K T2 = 250 K Q2 = 750 Q1 = ?

1 1

2 2

Q T

Q T=

1

300Q 750

250= ´ = 900 cal

Illustration 71.

The temperature inside and outside of refrigerator are 260 K and 315 K respectively. Assuming that the

refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done.

SolutionT2 = 260 K, T1 = 315 K ; W = 1 joule

Coefficient of performance of Carnot refrigerator 2 2

1 2

Q TW T T

b = =-

\ = =-

2Q 260 2601 315 260 55

Þ =2

260Q

55= 4.73 J So Q1 = Q2 + W = 5.73 J

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

70 E


A refrigerator takes heat from water at 0°C and transfer it to room at 27°C. If 100 kg of water is converted

in ice at 0°C then calculate the work done. (Latent heat of ice is 3.4 × 105 J/kg)

Solution

Coefficient of performance (COP) = 2

1 2

T 273 273T T 300 273 27

= =- -

52Q mL 100 3.4 10

WCOP COP 273/27

´ ´= = =

5100 3.4 10 27273

´ ´ ´= = 3.36 × 106 J

BEGINNER'S BOX-9

1. An ideal engine which works between 27°C & 227°C takes 100 calorie of heat in one cycle. Calculate the

work done in one cycle ?

2. The coefficient of performance of a carnot refrigerator is 5/3. If the refrigerator operating at a intake temperature

– 23°C then what will be the exhaust temperature ?

3. A carnot engine absorbs 1000 J of heat from a reservoir at 127°C and reject 600 J of heat during each

cycle.

Calculate (a) The efficiency of engine

(b) The temperature of the sink

(c) The amount of the useful work done during each cycle.

4. Two carnot engine A and B use one after another. Engine A absorb heat from source at T1 = 800 K and

release heat to sink of temp T2K. Second engine B absorb heat which release by first (at T2K) and release

heat at another sink at temperature T3 = 300 K. If work done by two engine is same than calculate value

of T2 ?

5. A carnot engine, efficiency 40% and temperature of sink 300 K to increase efficiency up to 60% calculate

change in temperature of source ?

6. An ideal refrigerator is working between temperature 27°C and 127°C. If it expells 120 calorie of heat in

one second then calculate its wattage ?

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

71E


BEGINNER'S BOX-1

1. 50 K < 50°F < 50°C

2. (a) All tie (b) 50°X, 50°Y, 50°W.

3. – 40°C or – 40°F.

4. (B) 5. 1.44 × 10–2 cm

6. In bimetallic strips the two metals have different

thermal expansion coefficient. Hence on heating

it bents towards the metal with lower thermal

expansion coefficient.

BEGINNER'S BOX-2

1. 191.11 Cal. 2. 50 kcal 3. 1 cal

4. 10° C 5. 1 : 1 6. 615 kcal.

7. 4000 cal 8. 160°C

9. 7.5°C, water 10. (A)

BEGINNER'S BOX-3

1.(a) The thermal conductivity of brass is high, i.e., brass

is a good conductor of heat. So, when a brass

tumbler is touched, heat quickly flows from human

body to tumbler. Consequently, the tumbler appears

colder. On the other hand, wood is a bad conductor.

So, heat does not flow from the human body to the

wooden tray in this case. Thus, it appears

comparatively hotter.

(b) This is because the two thin cloth enlose a layer of

air between them. Since air is a bad conductor of

heat therefore the conduction of heat is prevented.

(c) Because mudhouses are bad conductor of heat.

(d) Because air is trapped between their wings since

air is bad conductor of heat therefore the conduction

of heat is prevented.

(e) This is because wool contains air in its pores. Air

is a bad conductor of heat. In fact, wool is also a

bad conductor of heat so both the air and wool

donot permit heat to be conducted away from. So

woolen clothes are warm.

2. (a) 240 W (b) 1.5 3. 40°C

4. q1 = 116°C, q2 = 74°C

5. 68°C; 0.9 °C/cm 6. 7 : 2

BEGINNER'S BOX-4

1.(a) A body whose reflectivity is large would naturally absorb

less heat. So, a body with large reflectivity is a poor

absorber of heat. Poor absorbers are poor emitters.

So, a body with large reflectivity is a poor emitter.

(b) Let T be the temperature of the hot iron in the furnace.

Heat radiated per second per unit area, E = sT4

When the body is placed in the open at temperature

T0 , then the heat radiated/second/area, E' = s (T4 – T04)

Clearly E' < E. So, the optical pyrometer gives too

low a value for the temperature in the open.

(c) The lower layers of earth's atmosphere reflect

infrared radiations from earth back to the surface of

earth. Thus the heat radiation received by the earth

from the sun during the day are kept trapped by the

atmosphere. If atmosphere of earth were not there,

its surface would become too cold to live.

(d) When the electric heater is switched on, a stage is

quickly reached when the rate at which heat is

generated by electirc. Current becomes equal to the

rate at which heat is lost by radiation. This is case of

thermal equilibrium.

(e) Because black surface is good absorber of heat.

(f) On a clear night, the earth radiates energy into space

at a rate proportional to the fourth power of its

temperature (about 300K). The incoming radiation

from space is very small because its average

temperarure is nearly absolute zero. On the other

hand with cloud over, the earth radiates at 300K, but

the radiation is absorbed in the clouds, which radiate

energy back to earth again the radiation is trapped,

like the green house effect.

(g) This is due to the fact that a vaccum is created

between the two walls of the thermos flask. Heat can

enither flow from inside the flask to outside nor from

outside air to the liquid inside the flask. In this way,

loss of heat by conduction and convection has been

minimised. To minimise loss of heat by radiation the

surface is made shining.

2.2

1

rr 3. 40 °C 4. 28 °C

5. 1927 K

ANSWERS

Z:\N

OD

E02\

B0A

I-B0\

TARG

ET\P

HY\E

NG

\MO

DUL

E_03

\01-

THER

MA

L PH

YSIC

S\01

-TH

EORY

.P65

72 E


BEGINNER'S BOX-5

1.12

mole 2. 120 kPa 3. 2.2 × 1024

4. (a) 1

2

N 3N 2

= ; (b) 0.947

BEGINNER'S BOX-6

1. (a) 6.21 × 10–21 J; (b)10.35 × 10–21 J; (c) 6231J

2. (a) 127°C, (b) 527°C

3. (a) 3, (b) 53

, (c) 450R

4. vavg = 1 2 3v v v3

+ +,

Root mean square speed = 2 2 21 2 3v v v

3+ +

5. 9.2 m/s

6. (a) 1

2

E 1E 1

= (Q T = same for both);

(b) vrms = =70.9 1.3339.9 1

BEGINNER'S BOX-7

1. (a) WABCA = 12

× (2V0 – V0) × (3P0 – P0) = P0V0

(b) 52

P0V0 and 3P0V0 , (c) P V0 0

2

2. –5 × 105 J

3. WAB = 6000 J

WBC = zero

WCD = 1000 J

WDA = zero

4. (a) 24× 102 J; (b) 36× 102 J; (c) 60× 102 J

5. 4.16 J/cal

BEGINNER'S BOX-8

1. 3650 J

2. (a) 162.5 cal; (b) 237.5 cal

3. (A) – p, (B) – q , (C) – r

4. (A) – q , (B) – r , (C) – p , (D) – s

5. (A) –p, s, (B) –p, q, r, s, (C) –s ,(D) –q, s, (E) –r, s

6. (a) 700 J; (b) 500 J

7. (a) TB = 150 K ; TC = 37.5 K; (b) 2400 J

8. 1152.6 joule.

BEGINNER'S BOX-9

1. 168 J 2. 400 K

3. (a) 40%; (b) 240K; (c) 400 J

4. 550 K 5. 250 K 6. 126 watt.

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

73E

Pre-Medical : PhysicsALLENEXERCISE-I (Conceptual Questions) Build Up Your Understanding

TEMPERATURE & THERMAL EXPANSION1. At what temperature does the temperature in

Celsius and Fahrenheit equalise

(1) –40° (2) 40° (3) 36.6° (4) 38°

2. A difference of temperature of 25° C is equivalent

to a difference of :

(1) 45° F (2) 72° F (3) 32° F (4) 25° F

3. Which of the curves in figure represents the relation

between Celsius and Fahrenheit temperature?

(1) Curve a (2) Curve b

(3) Curve c (4) Curve d

C

c

F

a

b

d

4. The graph AB shown in figure is a plot of

temperature of a body in degree Celsius and

degree Fahrenheit. Then

32°F 212°F FahrenheitA

Cen

tigra

de

100°C B

(1) slope of line AB is 9/5




5. Oxygen boils at –183°C. This temperature is

approximately in Fahrenheit is :-

(1) –329°F (2) –261°F

(2) –215°F (4) –297°F

6. Using which of the following instrument, the

temperature of the sun can be determined ?

(1) Platinum thermometer

(2) Gas thermometer

(3) Pyrometer

(4) Vapour pressure thermometer

7. Two thermometers X and Y have ice points marked

at 15° and 25° and steam points marked as 75°

and 125° respectively. When thermometer X

measures the temperature of a bath as 60° on it,

what would thermometer Y read when it is used to

measure the temperature of the same bath ?

(1) 60° (2) 75° (3) 100° (4) 90°

8. The figure below shows four isotropic solids having

positive coefficient of thermal expansion. A student

predicts that on heating the solid following things

can happen. Mark true (T) or False (F) for comments

made by the student.

A

Ba

(i) The angle a in figure (1) will not change.

(ii) The length of line in figure (2) will decrease.

(iii) The radius of inner hole will decrease.

(iv) The distance AB will increase.

(1) T F F T (2) F T T F

(3) T T T T (4) F F T F

9. At STP a rod is hung from a frame as shown in

figure, leaving a small gap between the rod and

floor. The frame and rod system is heated uniformly

upto 350 K. Then

frame

Rod

(1) The rod will never touch the floor in any case.

(2) If arod > aframe, then rod may touch the floor.

(3) If arod < aframe, then rod may touch the floor.

(4) None of the above

10. A steel tape gives correct measurement at 20°C. A

piece of wood is being measured with the steel tape

at 0°C. The reading is 25 cm on the tape, the real

length of the given piece of wood must be:

(1) 25 cm (2) <25 cm

(3) >25 cm (4) can not say

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

74 E

Pre-Medical : Physics ALLEN11. The volume of a metal sphere increases by 0.15 %

when its temperature is raised by 24°C.The

coefficient of linear expansion of metal is :

(1) 2.5 × 10–5 /°C (2) 2.0 × 10–5 /°C

(3) –1.5 × 10–5 /°C (4) 1.2 × 10–5 /°C

12. Suppose there is a hole in a copper plate. On

heating the plate, diameter of hole, would :

(1) always increase

(2) always decrease

(3) always remain the same

(4) none of these

13. If two rods of length L and 2L having coefficients of

linear expansion a and 2a respectively are

connected so that total length becomes 3L, the

average coefficient of linear expansion of the

composition rod equals:

(1) 32

a (2) 52

a

(3) 53

a (4) none of these

14. The table gives the initial length l0, change in

temperature DT and change in length Dl of four

rods. Which rod has greatest coefficient of linear

expansion

Rod (m) T(°C) (m)l l0 D D

A1 1 100 1

A2 1 100 2

A3 1.5 50 3

A4 2.5 20 4

(1) A1 (2) A2 (3) A3 (4) A4

15. An iron bar (Young’s modulus = 1011 N/m2 ,

a = 10–6 /°C) 1 m long and 10–3 m2 in area is

heated from 0°C to 100°C without being allowed to

bend or expand. Find the compressive force

developed inside the bar.

(1) 10,000 N (2) 1000 N

(3) 5000 N (4) 105 N

16. A rod of length 2m rests on smooth horizontal floor.

If the rod is heated from 0°C to 20°C. Find the

longitudinal strain developed?

(a = 5 × 10–5/°C)

(1) 10–3 (2) 2 × 10–3 (3) Zero (4) None

CALORIMETRY

17. A body of mass 5 kg fal ls from a height of

30 metre. If its all mechanical energy is changed

into heat, then heat produced will be:-

(1) 350 cal (2) 150 cal

(3) 60 cal (4) 6 cal

18. A bullet moving with velocity v collides against wall.

consequently half of its kinetic energy is converted

into heat. If the whole heat is acquired by the bullet,

the rise in temperature will be:–

(1) v2/4S (2) 4v2 / 2S

(3) v2 / 2S (4) v2 / S

19. The amount of heat required in converting 1 g ice

at -100C into steam at 1000C will be :–

(1) 3028 J (2) 6056 J

(3) 721 J (4) 616 J

20. 2 kg ice at – 20°C is mixed with 5 kg water at

20°C. Then final amount of water in the mixture

would be;

Given specific heat of ice = 0.5 cal/g°C,

Specific heat of water = 1 cal/g°C,

Latent heat of fusion for ice = 80 cal/g.

(1) 6 kg (2) 5 kg (3) 4 kg (4) 2 kg

21. Two identical masses of 5 kg each fall on a wheel

from a height of 10m. The wheel disturbs a mass of

2kg water, the rise in temperature of water will

be :

(1) 2.6° C (2) 1.2° C (3) 0.32° C (4) 0.12° C

22. A block of mass 2.5 kg is heated to temperature of

500°C and placed on a large ice block. What is the

maximum amount of ice that can melt (approx.).

Specific heat for the body = 0.1 cal/g°C.

(1) 1 kg (2) 1.5 kg (3) 2 kg (4) 2.5 kg

23. 1 kg of ice at – 10°C is mixed with 4.4 kg of water

at 30°C. The final temperature of mixture is :

(specific heat of ice = 2100 J/kg-k)

(1) 2.3°C (2) 4.4°C (3) 5.3°C (4) 8.7°C

24. The amount of heat required to convert 1 g of ice at

0°C into steam at 100°C, is

(1) 716 cal. (2) 500 cal.

(3) 180 cal. (4) 100 cal.

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

75E

Pre-Medical : PhysicsALLEN25. The latent heat for vapourisation for 1 g water is

536 cal. Its value in Joule/kg will be :-

(1) 2.25 × 106 (2) 2.25 × 103

(3) 2.25 (4) None of these

26. If 10 g ice at 0°C is mixed with 10 g water at 20°C,

the final temperature will be :-

(1) 50°C (2) 10°C (3) 0°C (4) 15°C

27. 420 joule of energy supplied to 10 g of water will

raise its temperature by nearly :-

(1) 1°C (2) 4.2°C (3) 10°C (4) 42°C

28. A solid material is supplied with heat at a constant

rate. The temperature of material is changing with

heat input as shown in the figure. What does slope

DE represent.

(1) latent heat of liquid

(2) latent heat of vapour

(3) heat capacity of vapour

(4) inverse of heat capacity of vapour

29. The graph shown in the figure represent change in

the temperature of 5 kg of a substance as it abosrbs

heat at a constant rate of 42 kJ min–1. The latent

heat of vapourazation of the substance is :

Tem

pera

ture

(1) 630 kJ kg–1 (2) 126 kJ kg–1

(3) 84 kJ kg–1 (4) 12.6 kJ kg–1

30. A block of ice with mass m falls into a lake. After

impact, a mass of ice m/5 melts. Both the block of

ice and the lake have a temperature of 0°C. If L

represents the heat of fusion, the minimum distance

the ice fell before striking the surface is

(1) L5g (2)

5Lg (3)

gL5m

(4) mL5g

31. 10 g of ice at 0°C is kept in a calorimeter of water

equivalent 10 g. How much heat should be supplied

to the apparatus to evaporate the water thus

formed? (Neglect loss of heat)

(1) 6200 cal (2) 7200 cal

(3) 13600 cal (4) 8200 cal

32. Figure shows the temperature variation when heat

is added continuously to a specimen of ice (10 g) at

–40 °C at constant rate.

(Specific heat of ice = 0.53 cal/g °C and Lice = 80

cal/g, Lwater= 540 cal/g)

0

-40Q1 Q2 Q3 Q4

100

Q(cal)

Temp. (°C)

Column–I Column–II(A) Value of Q1 (in cal) (P) 800

(B) Value of Q2 (in cal) (Q) 1000

(C) Value of Q3 (in cal) (R) 5400

(D) Value of Q4 (in cal) (S) 212

(T) 900

(1) A®S; B®P; C®Q; D®T

(2) A®P; B®S; C®Q; D®R

(3) A®P; B®S; C®R; D®Q

(4) A®S; B®P; C®Q; D®R

33. The thermal capacity of any body is

(1) a measure of its capacity to absorb heat

(2) a measure of its capacity to provide heat

(3) the quantity of heat required to raise its

temperature by a unit degree

(4) the quantity of heat required to raise the

temperature of a unit mass of the body by a unit

degree

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

76 E

Pre-Medical : Physics ALLEN34. 2 litre water at 27°C is heated by a 1 kW heater in

an open container. On an average heat is lost to

surroundings at the rate 160 J/s. The time required

for the temperature to reach 77°C is

(1) 8 min 20 sec (2) 10 min

(3) 7 min (4) 14 min

35. A piece of ice (heat capacity = 2100 J kg–1 °C–1

and latent heat = 3.36 × 105 J kg–1) of mass m

grams is at –5°C at atmospheric pressure. It is given

420 J of heat so that the ice starts melting. Finally

when the ice-water mixture is in equilibrium, it is

found that 1 g of ice has melted. Assuming there is

no other heat exchange in the process, the value of

m is

(1) 40 g (2) 8 g (3) 16 g (4) 24 g

36. A 2100 W continuous flow geyser (instant geyser)

has water inlet temperature = 10°C while the water

flows out at the rate of 20 g/s. The outlet

temperature of water must be about

(1) 20°C (2) 30°C (3) 35°C (4) 40°C

37. A continuous flow water heater (geyser) has an

electrical power rating = 2 kW and efficienty of

conversion of electrical power into heat = 80%. If

water is flowing through the device at the rate of

100 cc/s, and the inlet temperature is 10°C, the

outlet temperature will be

(1) 12.2°C (2) 13.8°C

(3) 20°C (4) 16.5°C

CONDUCTION AND CONVECTION

38. The ratio of coefficient of thermal conductivity of

two different materials is 5:3. If the thermal

resistance of rods of same area of these material is

same, then what is ratio of length of these rods -

(1) 3:5 (2) 5:3 (3) 25:9 (4) 9:25

39. Rate of heat flow through a cylindrical rod is Q1.

Temperatures of ends of rod are T1 and T2. If all the

linear dimensions of the rod become double and

temperature difference remains same, it's rate of

heat flow is Q2, then :–

(1) Q1 = 2Q2 (2) Q2 = 2Q1

(3) Q2 = 4Q1 (4) Q1 = 4Q2

40. A heat flux of 4000 J/s is to be passed through a

copper rod of length 10 cm and area of cross section

100 cm2. The thermal conductivity of copper is 400

W/m°C. The two ends of this rod must be kept at a

temperature difference of

(1) 1°C (2) 10°C

(3) 100°C (4) 1000°C

41. The coefficient of thermal conductivity of copper is

nine times that of steel. In the composite cylindrical

bar shown in the figure what will be the temperature

at the junction of copper and steel ?

(1) 75°C 100°C

copper steel

0°C

18cm 6cm

(2) 67°C

(3) 33°C

(4) 25°C

42. A composite rod made of three rods of equal length

and cross-section as shown in the fig. The thermal

conductivities of the materials of the rods are K/2,

5K and K respectively. The end A and end B are at

constant temperatures. All heat entering the face

A goes out of the end B and there being no loss of

heat from the sides of the bar. The effective thermal

conductivity of the bar is

BA

K5KK/2

(1) 15K/16 (2) 6K/13

(3) 5K/16 (4) 2K/13.

43. The figure shows the face and interface

temperature of a composite slab containing of four

layers of two materials having identical thickness.

Under steady state condition, find the value of

temperature q.

(1) 5°C (2) 10°C

(3) –15°C (4) 15°C

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

77E

Pre-Medical : PhysicsALLEN44. Three rods made of the same material and having

the same cross-section have been joined as shown

in the figure. Each rod is of the same length. The

left and right ends are kept at 0°C and 90°C

respectively. The temperature of the junction of the

three rods will be :

(1) 45°C (2) 60°C (3) 30°C (4) 20°C

45. Two identical square rods of metal are welded end

to end as shown in figure (1), 20 calories of heat

flows through it in 4 minutes. If the rods are welded

as shown in figure (2), the same amount of heat will

flow through the rods in -

0°C

100°C0°C 100°C

l l

l

(b)(a)

(1) 1 minute (2) 2 minutes

(3) 4 minutes (4) 16 minutes

46. The coefficient of thermal conductivity depends

upon-

(1) Temperature difference of two ends

(2) Area of the plate

(3) Thickness of the plate

(4) Material of the plate

47. Which of the following cylindrical rods will conduct

most heat, when their ends are maintained at the

same steady temperature

(1) Length 1 m; radius 1 cm




48. Gravitational force is required for –

(1) Stirring of liquid (2) Convection

(3) Conduction (4) Radiation

49. The layers of atmosphere are heated through -

(1) Convection (2) Conduction

(3) Radiation (4) 2 and 3 both

50. The lengths and radii of two rods made of same

material are in the ratios 1 : 2 and 2 : 3 respectively.

If the temperature difference between the ends for

the two rods be the same then in the steady state.

The amount of heat flowing per second through

them will be in the ratio of

(1) 1 : 3 (2) 4 : 3

(3) 8 : 9 (4) 3 : 2

51. Two metal rods, 1 & 2 of same length have same

temperature difference between their ends, their

thermal conductivities are K1 & K2 and cross

sectional areas A1 & A2 respectively. What is

required condition for same rate of heat conduction

in them.

(1) K1 = K2 (2) K1 A1 = K2 A2

(3) 1 2

1 2

K K=

A A (4) 1 2

1 22 2

K K=

l l

52. The temperature of hot and cold end of a 20 cm

long rod in thermal steady state are at 100°C and

20°C respectively. Temperature at the centre of

the rod is

(1) 50°C (2) 60°C

(3) 40°C (4) 30°C

53. Consider a compound slab consisting of two different

materials having equal thicknesses and thermal

conductivities K and 2K, respectively. The equivalent

thermal conductivity of the slab is

(1) 3K (2) 43

K

(3) 23

K (4) 2 K

54. Under steady state, the temperature of a body

(1) Increases with time

(2) Decreases with time

(3) Does not change with time and is same at all the

points of the body

(4) Does not change with time but is different at

different points of the body

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

78 E

Pre-Medical : Physics ALLEN55. The area of the glass of a window of a room is

10m2 and thickness 2 mm. The outer and inner

temperature are 40°C and 20°C respectively.

Thermal conductivity of glass in MKS system is 0.2

then heat flowing in the room per second will be -

(1) 3 × 104 joules (2) 2 × 104 joules

(3) 30 joules (4) 45 joules

56. If the coefficient of conductivity of aluminium is 0.5

cal/cm-sec-°C, then in order to conduct

10 cal/sec-cm2 in the steady state, the temperature

gradient in aluminium must be

(1) 5°C/cm (2) 10°C/cm

(3) 20°C/cm (4) 10.5°C/cm

57. The dimensional formula for thermal resistance is

(1) M–1L–2T3q (2) M–1L–2T–3q

(3) ML2T–2q (4) ML2T2q–1

58. The material used in the manufacture of cooker

must have (K-coefficient of thermal conductivity,

S– specific heat of material used) :

(1) high K and low S

(2) low K and low S

(3) high K and high S

(4) low K and high S

59. The cause of air currents from ocean to ground is

example of

(1) The specific heat of water is more than that of

sand

(2) Convection

(3) Radiation

(4) Diffraction

60 On a cold morning, a person will feel metal surface

colder to touch than a wooden surface because

(1) Metal has high specific heat

(2) Metal has high thermal conductivity

(3) Metal has low specific heat

(4) Metal has low thermal conductivity

61. The ratio of the diameters of two metallic rods of

the same metarial is 2 : 1 and their lengths are in

the ratio 1 : 4. If the temperature difference between

them are equal, the rate of flow of heat in them will

be in the ratio of -

(1) 2 : 1 (2) 4 : 1 (3) 8 : 1 (4) 16 : 1

62. Two bars of thermal conductivities K and 3K and

lengths 1 cm and 2 cm respectively have equal

cross-sectional area. they are joined length wise as

shown in the figure. If the temperature at the ends

of this composite bar is 0°C and 100°C respectively

(see figure), then the temperature of the interface

(f) is

K 3K

Fig.1 cm

0°C

f

2 cm

100°C

(1) 50°C (2) 100

3°C (3) 60°C (4)

2003

°C

63. Mud houses are cooler in summer and warmer in

winter because

(1) Mud is super conductor of heat

(2) Mud is good conductor of heat

(3) Mud is bad conductor of heat

(4) None of these

64. Two walls of thicknesses d1 and d2 and thermal

conductivity k1 and k2 are in contact. In the steady

state, if the temperatures at the outer surface are

T1 and T2, the temperature at the common wall is-

(1) 1 1 2 2 2 1

1 2 2 1

K T d K T dK d K d

++ (2)

1 1 2 2

1 2

K T K Td d

++

(3) 1 1 2 2

1 21 2

K d K dT T

T T

æ ö+ç ÷+è ø

(4) 1 1 1 2 2 2

1 1 2 2

K d T K d TK d K d

++

65. In which of the following phenomenon heat

convection does not take place

(1) land and sea breeze

(2) boiling of water

(3) heating of glass surface due to filament of the

bulb

(4) air around the furance

66. In natural convection, a heated portion of a liquid

moves because :

(1) Its molecular motion becomes aligned

(2) Of moleuclar collisions within it

(3) Its density is less than that of the surrounding

fluid

(4) Of currents of the surrounding fluid

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

79E

Pre-Medical : PhysicsALLEN67. It is hotter at the same distance over the top of a

fire than it is in the side of it, mainly because

(1) Air conducts heat upwards

(2) Heat is radiated upwards

(3) Convection takes more heat upwards

(4) Convection, conduction and radiation all

contribute significantly transferring heat upward

RADIATION

68. A spherical body of area A, and emissivity e = 0.6

is kept inside a black body. What is the rate at which

energy is radiated per second at temperature T.

(1) 0.6 s AT4 (2) 0.4 s AT4

(3) 0.8 s AT4 (4) 1.0 s AT4

69. Radius of two spheres of same material are

1 & 4 m respectively and their temperature are

4´103 and 2´103 K respectively. Then ratio of

emitted energy of spheres per sec. will be -

(1) 1:2 (2) 2:1 (3) 1:1 (4) 4:1

70. Cooling rate of a sphere of 600 K at external

environment (200 K) is R. When the temperature

of sphere is reduced to 400 K then cooling rate of

the sphere becomes :

(1) 3

16 R (2)

163

R (3) 927

R (4) None

71. Temperature of a body is 4000C. Assuming the

surrounding temperature to be negligible. At what

temperature will body emit double energy radiation?

(1) 2000 C (2) 200 K

(3) 8000 C (4) 800 K

72. If temperature of ideal black body increased by

10%, then percentage increase in quantity of

radiation emitted from it's surface will be :-

(1) 10% (2) 40% (3) 46% (4) 100%

73. The rectangular surface of area 8cm × 4 cm of a

black body at a temperature of 127°C emits energy

at the rate of E. If the length and breadth of the

surface are each reduced to half of the initial value

and the temperature is raised to 327°C, the rate of

emission of energy will become.

(1) 3

E8

(2) 81

E16

(3) 9

E16

(4) 81

E64

74. The rate of emission of radiation of a black body at

273°C is E, then the rate of emission of radiation

of this body at 0°C will be

(1) E

16(2)

E4

(3) E8

(4) 0

75. If a liquid takes 30 s in cooling from 95°C to 90°C

and 70 s in cooling from 55°C to 50°C then

temperature of room is -

(1) 16.5°C (2) 22.5°C

(3) 28.5°C (4) 32.5°C

76. The thermal capacities of two bodies are in the ratio

of 1:4. If the rate of loss of heat are equal for the

two bodies under identical conditions of

surroundings, then the ratio of rate of fall of

temperature of the two bodies is -

(1) 1:4 (2) 4:1

(3) 1:8 (4) 8:1

77. Newton's law of cooling is used in laboratory for the

determination of the

(1) Specific heat of the gases

(2) The latent heat of gases

(3) Specific heat of liquids

(4) Latent heat of liquids

78. The um – T curve for a perfect black body is –

(um ® frequency corresponding to maximum

emission of radiation)

DBC

A

um

T

(1) A (2) B (3) C (4) D

79. Two stars appear to be red and blue, what is true

about them -

(1) The red star is nearer

(2) The blue star is nearer

(3) The temperature of red star is more

(4) The temperature of blue star is more

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

80 E

Pre-Medical : Physics ALLEN80. The temperature of a furnace is 23240C and the

intensity is maximum in its radiation spectrum nearly

at 12000 A0. If the intensity in the spectrum of a

star is maximum nearly at 4800 A0, then the sur-

face temperature of star is

(1) 84000C (2) 72000C

(3) 6219.50C (4) 59000C

81. There is a black spot on a body. If the body is heated

and carried in a dark room then it glows more. This

can be explained on the basis of –

(1) Newton's law of cooling

(2) Wein's law

(3) Kirchhoff's law

(4) Stefan's law

82. The colour of a star is an indication of its -

(1) Weight (2) Distance

(3) Temperature (4) Size

83. If a carved black utensil is heated to high

temperature and then brought in dark then :

(1) Both utensil and its carving will shine

(2) Only carving will shine

(3) Only utensil will shine

(4) None of the utensil and carving will shine

84. According to Newton's law of cooling, the rate of

cooling of a body is proportional to :–

(1) Temperature of the body

(2) Temperature of the surrounding

(3) Fourth power of the temperature of body

(4) Difference of the temperature of the body and

the surrounding.

85. The original temperature of a black body is 727°C.

Calculate temperature at which total radiant energy

from this black body becomes double :

(1) 971 K (2) 1190 K

(3) 2001 K (4) 1458 K

86. Calculate the energy radiated per minute from the

filament of an incandascent lamp at 2000K if the

surface area is 5 × 10–5 m2 and its relative emittance

is 0.85, s = 5.7 × 10–8 MKS units :

(1) 1230 J (2) 2215 J

(3) 2115 J (4) 2325 J

87. Ratio of radius of curvature of cylindrical emitters

of same material is 1:4 and their temperature are

in ratio 2:1. Then ratio of amount of heat emitted

by them is - (For Cylinder length = radius)

(1) 2:1 (2) 1:1

(3) 4:1 (4) 1:4

88. The ideal black body is :

(1) Hot coal at high temperature

(2) Surface of glass printed with coaltar

(3) Metal surface

(4) A hollow container painted with black colour

89. The energy emitted per second by a black body at

270C is 10 J. If temperature of the black body is

increased to 3270C , the energy emitted per second

will be :-

(1) 20 J (2) 40 J

(3) 80 J (4) 160 J

90. Energy is being emitted from the surface of black

body at 127°C at the rate of 1.0 × 106 J/s m2. The

temperature of black body at which the rate of

energy is 16.0 × 106 J/s m2 will be :

(1) 754°C (2) 527°C

(3) 254°C (4) 508°C

91. Solar constant for earth is 2 cal/min cm2, if distance

of mercury from sun is 0.4 times than distance of

earth from sun then solar constant for mercury will

be?

(1) 12.5 cal/min cm2

(2) 25 cal/min cm2

(3) 0.32 cal/min cm2

(4) 2 cal/min cm2

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

81E

Pre-Medical : PhysicsALLEN92. Two spherical bodies A (radius 6 cm) and B (radius

18 cm) are at temperature T1 and T2 respectively.

The maximum intensity in the emission spectrum

of A is at 500 nm and in that of B is at 1500 nm.

Considering them to be black bodies, what will be

the ratio of the rate of total energy radiated by A to

that of B ?

(1) 9 (2) 6 (3) 12 (4) 3

93. Star S1 emits maximum radiation of wavelength

420 nm and the star S2 emits maximum radiation

of wavelength 560 nm, what is the ratio of the

temperature of S1 and S2 :

(1) 4/3 (2) (4/3)1/4

(3) 3/4 (4) (3/4)1/2

94. If a piece of metal is heated to temperature q and

then allowed to cool in a room which is at

temperature q0, the graph between the

temperature T of the metal and time t will be closest

to

(1)

tO

q0

T

(2)

tO

q0

T

(3) (4)

95. A liquid in a beaker has temperature q at time t

and q0 is temperature of surroundings, then

according to Newton's law of cooling, correct graph

between loge(q – q0) and t is:

(1)

t

log

( –

)

e0

qq

(2)

t

log

( –

)

e0

qq

(3)

t

log

( –

)

e0

qq

0

(4)

t

log

( –

)

e0

qq

0

96. A bucket full of hot water cools from 750C to 700C

in time T1, from 700 to 650C in time T2 and from

650C to 600C in time T3, then

(1) T1 = T2 = T3 (2) T1 > T2 > T3

(3) T1 < T2 < T3 (4) T1 > T2 < T3

97. The Wein's displacement law express relation

between :-

(1) Wavelength corresponding to maximum energy

and temperature.

(2) Radiation energy and wavelength

(3) Temperature and wavelength

(4) Colour of light and temperature

98. Four indentical calorimeters painted in different

colours, are heated to same temperature and then

allowed to cool in vacuum. Which will cool

fastest ?

(1) One which is painted bright

(2) One which is painted thick white

(3) One which is painted thick black

(4) One which is painted bright white

99. A body cools from 60°C to 50°C in 10 minutes. If

the room temperature is 25°C and assuming

Newotn's cooling law holds good, the temperature

of the body at the end of next 10 minutes is :

(1) 45°C (2) 42.85°C

(3) 40°C (4) 38.5°C

100. As compared to the person with white skin, the

person with black skin will experience

(1) Less heat and more cold

(2) More heat and more cold

(3) More heat and less cold

(4) Less heat and less cold

101. We consider the radiation emitted by the human

body. Which of the following statements is true ?

(1) The radiation is emitted during the summers and

absorbed during the winters

(2) The radiation emitted lies in the ultraviolet region

and hence is not visible

(3) The radiation emitted is in the infra-red region

(4) The radiation is emitted only during the day

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

82 E

Pre-Medical : Physics ALLEN102. Shown below are the black body radiation curves

at temperatures T1 and T2 (T2 > T1). Which of the

following plots is correct :–

(1) (2)

(3) (4)

103. The radii of two spheres made of same metal are r

and 2r. These are heated to the same temperature

and placed in the same surrounding. The ratio of

rates of decrease of their temperature will be

(1) 1 : 1 (2) 4 : 1 (3) 1 : 4 (4) 2 : 1

104. If E is the total energy emitted by a body at a

temperature T K and Elmax is the maximum spectral

energy emitted by it at the same temperature, then-

(1) E µ T4; Elmax µ T5 (2) E µ T4; Elmax µ T–5

(3) E µ T–4; Elmax µ T4 (4) E µ T5; Elmax µ T4

105. If el and al be the emissive power and absorption

power respectively of a body and El be the emissive

power of an ideal black body, then from Kirchhoff's

laws

(1) al = El / el (2) al / el = El

(3) el / al = El (4) el = El / al

106. A liquid takes 5 min. to cool from 80°C to 50°C.

How much time it will take to cool from 60°C to

30°C. Temperature of surroundings is 20°C –

(1) 15 min. (2) 20 min. (3) 100 min. (4) 9 min.

107. A cup of tea cools from 80°C to 60°C in one minute.

The ambient temperature is 30°C. In cooling from

60°C to 50°C, it will take :

(1) 50 sec. (2) 90 sec (3) 60 sec. (4)48 sec.

108. If lm denotes the wavelength at which the radiation

emission from a black body at a temperature T K is

maximum then :

(1) lm a T (2) lm a T2

(3) lm a T-1 (4) lm a T-2

109. A body is in thermal equillbrium with the surrounding

:

(1) It will stop emitting heat radiation

(2) Amount of radiation emitted and absorbed by it

will be equal

(3) It will emit heat radiation at faster rate

(4) It will emit heat radiation slowly

110. Which of the following statement is correct for ideal

black body :

(1) This absorbs visible radiation only.

(2) This absorbs infrared radiation only

(3) This absorbs half of radiation only and reflects

the half

(4) This totally absorbs heat radiation of all the

wavelengths

111. Two spheres P and Q of same colour having radii 8

cm and 2 cm are maintained at temperatures 127°C

and 527°C respectively. The ratio of energy radiated

by P and Q is –

(1) 0.054 (2) 0.0034

(3) 1 (4) 2

112. On increasing the temperature of a black body,

wavelength for maximum emission.

(1) Shifts towards smaller wavelength

(2) Shifts towards greater wavelength

(3) Does not shift

(4) Depends on the shape of source.

113. A solid cube and sphere are made of same

substance and both have same surface area. If the

temperature of both bodies 1200 C then :

(1) Both will loss of Heat by same rate

(2) Rate of loss of Heat of cube will be more than

that of the sphere

(3) Rate of loss of Heat of the sphere will be more

than that of the cube

(4) Rate of loss of Heat will be more for that whose

mass is more

114. Two spheres of radii in the ratio 1 : 2 and densities

in the ratio 2 : 1 and of same specific heat, are

heated to same temperature and left in the same

surrounding. Their rate of falling temperature will

be in the ratio :

(1) 2 :1 (2) 1 : 1

(3) 1 : 2 (4) 1 : 4

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

83E

Pre-Medical : PhysicsALLEN115. A black body at 200 K is found to emit maximum

energy at a wavelength 14 mm. When its

temperature is raised to 1000 K, then wavelength

at which maximum energy emitted is :

(1) 14 mm (2) 15 mm

(3) 2.8 mm (4) 28 mm

116. The spectrum from a black body radiation is a :

(1) line spectrum

(2) band spectrum

(3) continuous spectrum

(4) line & band both

117. The temperature of a perfect black body is

727°C and its area is 0.1 m2. If stefan's constant

is 5.67 × 10–8 watt/m2–K4, then heat radiated

by it in 1 minute is :

(1) 8100 cal (2) 81000 cal

(3) 810 cal (4) 81 cal

118. A black body radiates energy at the rate of

E watt/m2 at a high temperature T K. When the

temperature is reduced to T2

K, the radiant energy

will be

(1) E

16(2)

E4

(3) 4 E (4) 16 E

119. In solar spectrum fraunhoffer's lines are presents

because :

(1) Definite absorption takes place in photosphere

of sun.

(2) Definite absorption takes place in chromosphere

of sun.

(3) These wave lengths are not at all emitted by

sun.

(4) Nuclear reactions take place in sun.

120. On reducing temperature of surface to one third,

amount of radiation becomes :-

(1) 127

(2) 181

(3) 19

(4) 13

121. The absorptive power of a perfectly black body is

(1) zero (2) infinity

(3) 1.5 (4) 1.0

KTG & GAS LAWS AND IDEAL GAS EQUATION

122. Find the approximate number of molecules

contained in a vessel of volume 7 litres at 0°C at

1.3 × 105 pascals

(1) 2.4 × 1023 (2) 3 × 1023

(3) 6 × 1023 (4) 4.8 × 1023

123. A real gas behaves like an ideal gas if its

(1) pressure and temperature are both high

(2) pressure and temperature are both low

(3) pressure is high and temperature is low

(4) pressure is low and temperature is high

124. The lowest pressure (the best vaccum) that can be

created in laboratory at 27°C is 10–11 mm of Hg.

At this pressure, the number of ideal gas molecules

per cm3 will be :-

(1) 3.22 × 1012 (2) 1.61 × 1012

(3) 3.21 × 106 (4) 3.22 × 105

125. Two gases of equal molar amount are in thermal

equilibrium. If Pa, Pb and Va, Vb are their respective

pressures and volumes, then which relation is true:–

(1) Pa ¹ Pb, Va = Vb

(2) Va = Vb, Va ¹ Vb

(3) Pa/Vb = Pb/Vb

(4) PaVa = PbVb

126. Equal volume of H2, O2 and He gases are at same

temperature and pressure. Which of these will have

large number of molecules :-

(1) H2

(2) O2

(3) He

(4) All the gases will have same number of molecules

127. Gases obey vander - waal's equation at :

(1) Only normal temperature and pressure

(2) Only high temperature and high perssure

(3) Only high temperature and low pressure

(4) All temperature and pressure

128. A box contains N molecules of a gas. If the

number of molecules is doubled, then the pressure

will :-

(1) Decrease (2) Be same

(3) Be doubled (4) Get tripled

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

84 E

Pre-Medical : Physics ALLEN129. An ideal gas mixture filled inside a balloon expands

according to the relation PV2/3 = constant. The

temperature inside the balloon is

(1) increasing (2) decreasing

(3) constant (4) can’t be said

130. A rigid tank contains 35 kg of nitrogen at 6 atm.

Sufficient quantity of oxygen is supplied to increase

the pressure to 9 atm, while the temperature

remains constant. Amount of oxygen supplied to

the tank is :

(1) 5 kg (2) 10 kg

(3) 20 kg (4) 40 kg

131. An ideal gas follows a process PT = constant. The

correct graph between pressure & volume is :-

(1) (2)

(3) (4)

132. A cyclic process ABCA is shown in P–T diagram.

When presented on P–V, it would

(1) (2)

(3) (4)

133. 28 g of N2 gas is contained in a flask at a pressure

10 atm. and at a temperature of 57°C. It is found

that due to leakage in the flask, the pressure is

reduced to half and the temperature reduced to

27°C. The quantity of N2 gas that leaked out is –

(1) 11/20 g (2) 80/11 g

(3) 5/63 g (4) 63/5 g

134. When temperature of a gas, contained in closed

vessel is increased by 50C, its pressure increases by

1%. The original temperature of the gas was

approximately :

(1) 5000C (2) 2730C

(3) 2270C (4) 500C

135. During an experiment an ideal gas obeys an addition

equation of state P2V = constant. The initial

temperature and volume of gas are T and V

respectively. When it expands to volume 2V, then

its temperature will be :

(1) T (2) 2 T

(3) 2 T (4) 2 2 T

136. 250 litre of an ideal gas is heated at constant

pressure from 270C such that its volume becomes

500 liters. The final temperature is :

(1) 540C (2) 3000C

(3) 3270C (4) 6000C

137. A balloon contains 500 m3 of helium at 270C and

1 atmosphere pressure. The volume of the helium

at –30C temperature and 0.5 atmosphere

pressure will be-

(1) 500 m3 (2) 700 m3

(3) 900 m3 (4) 1000 m3

138. A vessel has 6g of oxygen at pressure P and

temperature 400 K. A small hole is made in it so

that oxygen leaks out. How much oxygen leaks out

if the final pressure is P/2 and temperature is

300 K ?

(1) 3g (2) 2g (3) 4g (4) 5g

139. Relation PV = RT is given for following condition

for real gas -

(1) High temperature and high density

(2) Low temperature and low density

(3) High temperature and low density

(4) Low temperature and high density

140. A container of 5 litre has a gas at pressure of

0.8 m column of Hg. This is joined to an

evacuated container of 3 litre capacity. The

resulting pressure will be :- (At constant temp.)

(1) 4/3 (2) 0 .5 m

(3) 2 . 0 m (4) 3/4 m

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

85E

Pre-Medical : PhysicsALLEN141. At a given temperature, the pressure of an ideal

gas of density r is proportional to -

(1) 2

1r (2)

1r (3) r2 (4) r

142. O2 gas is filled in a cylinder. When pressure is

increased 2 times, temperature becomes four times.

then how much times their density will become :

(1) 2 (2) 4 (3) 14

(4) 12

143. On increasing the temperature of a gas filled in a

closed container by 10C its pressure increases by

0.4%, then initial temperature of the gas is-

(1) 250C (2) 2500C

(3) 250 K (4) 25000C

144. The variation of PV graph with V of a fixed mass of

a ideal gas at constant temperature is graphically

represented as shown in figure :

(1) PV

V

(2) PV

V

(3) PV

V

(4) PV

V

145. The number of oxygen molecules in a cylinder

of volume 1 m3 at a temperature of 27°C and

pressure 13.8 Pa is :

(Boltzmaan's constant k = 1.38 × 10–23 JK–1)

(1) 6.23 × 1026 (2) 0.33 × 1028

(3) 3.3 × 1021 (4) None of these

146. A cylinder contains 10 kg of gas at pressure of

107 N/m2. When final pressure is reduce to

2.5 × 106 N/m2 then quantity of gas taken out of

the cylinder will be : (temperature of gas is constant)

(1) 15.2 kg (2) 3.7 kg

(3) zero (4) 7.5 kg

147. Hydrogen and helium gases of volume V at same

temperature T and same pressure P are mixed to

have same volume V. The resulting pressure of the

mixture will be :

(1) P/2

(2) P

(3) 2P

(4) Depending on the relative mass of the gases

148. The equation of state for 5g of oxygen at a pressure

P and temperature T occupying a volume V, will be :–

(where R is the gas constant)

(1) PV = 5 RT

(2) PV = (5/2) RT

(3) PV = (5/16) RT

(4) PV = (5/32)RT

149. In kinetic theory of gases, it is assumed that

molecules :-

(1) Have same mass but can have different volume

(2) Have same volume but masses can be different

(3) Have both mass and volume different

(4) Have same mass but negligible volume

150. The volume of an ideal gas is V at pressure P and

temperature T . The mass of each molecule of the

gas in m. The density of gas will be :-

(K is Boltzmann's constant)

(1) mKT (2) Pm / KT

(3) P / KTV (4) P / KT

151. The thermodynamic variables of a jar filled with gas

A are P, V and T and another jar B filled with another

gas are 2P, V/4 and 2T, where the symbols have

their usual meaning. The ratio of the number of

molecules of jar A to those of jar B is :

(1) 4 : 1 (2) 2 : 1 (3) 1 : 2 (4) 1 : 1

152. At N.T.P. volume of a gas is changed to one fourth

volume, at constant temperature then the new

pressure will be :

(1) 2 atm. (2) 5 32 atm.

(3) 4 atm. (4) 1 atm.

153. A gas contained in a box of 0.1 m3 at atmospheric

pressure is connected to another vessel of

0.09 m3. Consequent change in pressure is X mm

of Hg. Then X in metre is -

(1) 0.4 (2) 0.5 (3) 0.36 (4) 0.3

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

86 E

Pre-Medical : Physics ALLEN154. Find the correct relation in given P-V diagram :

(1) T1 = T2

V

PT2

T1

(2) T1 > T2

(3) T1 < T2

(4) T1 T2

155. Simple behaviour under all conditions of real gas is

governed by the equation :-

(1) PV = µRT

(2) 2

aP (V b) RT

Væ ö+ - =ç ÷è ø

(3) PV = constant

(4) PVg = constant

VARIOUS SPEEDS, DEGREE OF FREEDOM,SPECIFIC HEAT CAPACITIES OF GASES ANDMEAN FREE PATH

156. Three particles have speeds of 2u , 10u and 11u.

Which of the following statements is correct?

(1) The r.m.s. speed exceeds the mean speed by

about u.

(2) The mean speed exceeds the r.m.s. speed by

about u.

(3) The r.m.s. speed equals the mean speed.

(4) The r.m.s. speed exceeds the mean speed by

more than 2u.

157. The root mean square velocity of the molecules of

an ideal gas is :-

(1) RT Mw/ (2) 3RT Mw/

(3) 3RTMw (4) RT Mw/ 3

158. At constant pressure hydrogen is having temperature

of 327° C. Till what temperature it is to be cooled so

that the rms velocity of its molecules becomes half of

the earlier value :-

(1) –123°C (2) 123°C (3) –100°C (4) 0°C

159. The rms velocity of gas molecules of a given amount

of a gas at 27°C and 1.0 × 105 N m-2 pressure is

200 m sec-1. If temperature and pressure are

respectively 127°C and 0.5 × 105 N m–2, the rms

velocity will be :-

(1) 400/ 3 ms-1 (2) 100 2 ms-1

(3) 100 2 /3 ms-1 (4) 50 23

ms-1

160. Two containers of same volume are filled with atomic

Hydrogen and Helium respectively at 1 and 2 atm

pressure. If the temperature of both specimen are

same then average speed < CH > for hydrogen

atoms will be -

(1) HC = 2 HeC (2) HC = HeC

(3) CH = 2 HeC (4) HC = HeC

2

161. The r.m.s. speed of a gas molecule is 300 m/s.

Calculate the r.m.s. speed if the molecular weight

is doubled while the temperature is halved-

(1) 300 m/s (2) 150 m/s

(3) 600 m/s (4) 75 m/s

162. The root mean square velocity of hydrogen

molecules at 300 K is 1930 m/s. Then the r.m.s.

velocity of oxygen molecules at 1200 K will be :

(1) 482.5 m/s

(2) 965.0 m/s

(3) 1930 m/s

(4) 3860 m/s

163. The rms velocity of H2 is 2 × 103 m /s. What will be

the rms velocity of O2 molecules at the same tem-

perature :-

(1) 103 m/s

(2) 500 m/s

(3) 0.5 × 104 m/s

(4) 3 × 103 m / s

164. For the molecules of an Ideal gas, Which of the

following velocity average can not be zero

(1) < v > (2) < v4 >

(3) < v3 > (4) < v5 >

165. The temperature at which root mean square velocity

of molecules of helium is equal to root mean square

velocity of hydrogen at N.T.P is-

(1) 2730C (2) 273 K

(3) 5460C (4) 844 K

166. If the pressure of a gas is doubled at constant

temperature, then the mean square velocity will

become :-

(1) No change (2) double

(3) Four times (4) None of the above

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

87E

Pre-Medical : PhysicsALLEN167. The reason for the absence of atmosphere on moon

is that the :

(1) Value of vrms of the molecules of gas is more

than the value of escape velocity

(2) Value of vrms of gas is less than escape velocity

(3) Value of vrms is negiligible


168. The speeds of 5 molecules of a gas (in arbitary

units) are as follows 2,3,4,5,6 The root mean square

speed for these moecules is -

(1) 2.91 (2) 3.52 (3) 4.00 (4) 4.24

169. The root mean square speed of the molecules of a

gas is :

(1) Independent of its pressure but directly

proportional to its Kelvin temperature

(2) Directly proportional to the square roots of both

its pressure and its Kelvin temperature

(3) Independent of its pressure but directly

proportional to the square root of its Kelvin

temperature

(4) Directly proportional to both its pressure and

its Kelvin temperature

170. At 00C temperature root mean square speed of

which of the following gases will be maximum:-

(1) H2 (2) N2 (3) O2 (4) SO2

171. The root mean square speed of hydrogen molecules

of an ideal hydrogen gas kept in a gas chamber is

3180 m/s. The pressure of the hydrogen gas is :-

(Density of hydrogen gas = 8.99 x 10-2 Kg/m3,

1 atmosphere = 1 .01 × 105 N/m2)

(1) 1.0 atmosphere (2) 1.5 atomsphere

(3) 2.0 atmosphere (4) 3.0 atomsphere

172. If the rms velocity of molecules of a gas in a

container is doubled then the pressure will:-

(1) Become four times (2) Also get doubled

(3) Be same (4) Become one half

173. The root mean square velocity of a gas molecule

of mass m at a given temperature is proportional

to –

(1) m0 (2) m

(3) m (4) 1

m

174. vrms, vav and vmp are root mean square, average and

most probable speeds of molecules of a gas obeying

Maxwell's velocity distribution. Which of the following

statements is correct

(1) vrms < vav < vmp (2) vrms > vav > vmp

(3) vmp < vrms < vav (4) vmp > vrms > vav

175. If the r.m.s. velocity of hydrogen becomes equal to

the escape velocity from the earth surface, then

the temperature of hydrogen gas would be-

(1) 1060 K (2) 5030 K

(3) 8270 K (4) 104 K

176. The pressure exerted by a gas in P0. If the mass of

molecules becomes half and their velocities become

double, then pressure will become

(1) 0P2

(2) P0 (3) 2P0 (4) 4P0

177. The root mean square (rms) speed of oxygen

molecules O2 at a certain temperature T (absolute)

is v. If the temperature is doubled and oxygen gas

dissociates into atomic oxygen. The rms speed :

(1) becomes v/ 2 (2) remains v

(3) becomes 2v (4) becomes 2v

178. If the root mean square speed of hydrogen

molecules is equal to root mean square speed of

oxygen molecules at 470C, the temperature of

hydrogen is -

(1) 20 K (2) 47 K

(3) 50 K (4) 94 K

179. The root mean square and most probable speed of

the molecules in a gas are :

(1) same

(2) different

(3) cannot say

(4) depends on nature of the gas

180. According to Maxwell's law of distribution of

velocites of molecules, the most probable velocity

is :-

(1) greater than the mean speed

(2) equal to the mean speed

(3) equal to the root mean square speed

(4) less than the root mean square speed

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

88 E

Pre-Medical : Physics ALLEN181. The ratio of average translational kinetic energy to

rotational kinetic energy of a diatomic molecule at

temperature T is

(1) 3 (2) 7/5

(3) 5/3 (4) 3/2

182. For hydrogen gas cP – cV = a and for oxygen gas

cP – cV = b then the relation between a and b is

(where cP & cV are gram specific heats)

(1) a = 16 b (2) b = 16 a

(3) a = b (4) None of these

183. A gas mixture consists of 2 moles of oxygen and 4

moles of argon at temperature T. Neglecting all

vibrational modes, the total internal energy of the

system is

(1) 4 RT (2) 15 RT

(3) 9 RT (4) 11 RT

184. The average kinetic energy of a gas molecule at

270C is 6.21 × 10-21 J. Its average kinetic energy at

2270C will be

(1) 52.2 × 10-21 J

(2) 5.22 × 10-21 J

(3) 10.35 × 10-21 J

(4) 11.35 × 10-21 J

185. Two containers A and B contain molecular gas at same

temperature with masses of molecules are mA and mB,

then relation of momentum PA and PB will be-

(1) PA = PB

(2) PA =

1/2

A

B

mm

æ öç ÷è ø

PB

(3) PA = mm

B

A

FHG

IKJ

1 2/

PB

(4) PA = A

B

mm

æ öç ÷è ø

PB

186. A cylinder of 200 litre capacity is containing H2.

The total translational kinetic energy of molecules

is 1.52 × 105 J.The pressure of H2 in the cylinder

will be in N m-2 :-

(1) 2 × 105 (2) 3 × 105

(3) 4 × 105 (4) 5 × 105

(187–188) Five moles of helium are mixed withtwo moles of hydrogen to form a mixture. Takemolar mass of helium M1 = 4g and that ofhydrogen M2 = 2g

187. The equivalent molar mass of the mixture is

(1) 6g (2) 13g

7(3)

18g7

(4) 24g7

188. The equivalent value of g in the above question is

(1) 1.59 (2) 1.53

(3) 1.56 (4) none

189. Two monoatomic ideal gas at temperature T1 and

T2 are mixed. There is no loss of energy. If the

mass of molecules of the two gases are m1 and m2

and number of their molecules are n1 and n2

respectively, then temperature of the mixture will

be :

(1) 1 2

1 2

T Tn n

++ (2)

1 2

1 2

T Tn n

+

(3) 2 1 1 2

1 2

n T n Tn n

++ (4)

1 1 2 2

1 2

n T n Tn n

++

190. The total kinetic energy of 1 mole of N2 at 270C

will be approximately :-

(1) 1500 J

(2) 1500 calorie

(3) 1500 kilo calorie

(4) 1500 erg.

191. Mean kinetic energy (or average energy) per gm.

molecule of a monoatomic gas is given by :

(1) 3RT/2 (2) kT/2

(3) RT/3 (4) 3kT/2

192. Relation between the ratio of specific heats (g) of

gas and degree of freedom 'f' will be

(1) g = f + 2 (2) 1 1 1

f 2= +

g

(3) f = 2 / (g-1) (4) f = 2( g-1)

193. Relation between pressure (P) and energy density

(E) of an ideal gas is -

(1) P = 2E

3(2) P =

3E

2

(3) P = 3E

5(4) P = E

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

89E

Pre-Medical : PhysicsALLEN194. On mixing 1 g mole of a monoatomic with 1 g mole

of a diatomic gas the specific heat of mixture at

constant volume will be :–

(1) R (2) 3/2 R (3) 2R (4) 5/2R

195. Absolute zero temperature is one at which-

(1) All liquids convert into solid

(2) All gases convert to solid

(3) All matter is in solid state

(4) The K.E. of molecules becomes zero

196. For a gas V

RC

= 0.67. This gas is made up of

molecules which are :

(1) Diatomic

(2) Mixture of diatomic and polyatomic molecules

(3) Monoatomic

(4) Polyatomic

197. If the total number of H2 molecules is double that

of the O2 molecules then ratio of total kinetic

energies of H2 to that of O2 at 300 K is :

(1) 1: 1 (2) 1 : 2

(3) 2 : 1 (4) 1 : 16

198. At which of the following temperature any gas has

average moleculer kinetic energy double that of at

200C

(1) 400C (2) 800C

(3) 3130C (4) 5860C

199. When temperature is increased from 0°C to 273°C,

in what ratio of final to initial the average kinetic

energy of molecules change ?

(1) 1 (2) 3 (3) 4 (4) 2

200. The kinetic energy associated with per degree of

freedom of a molecule is -

(1)12

MC2rms (2) kT

(3) kT / 2 (4) 3 kT/2

201. If 2 gm moles of a diatomic gas and 1 gm mole of a

mono-atomic gas are mixed then the value of g (Cp/

Cv) for mixture will be :–

(1) 1319

(2) 1913

(3) 75

(4) 53

202. Which of the following statement is true according

to kinetic theory of gases ?

(1) The collision between two molecules is inelastic

and the time between two collisions is less than

the time taken during the collision.

(2) There is a force of attraction between the

molecules

(3) All the molecules of a gas move with same

velocity

(4) The average of the distances travelled between

two successive collisions is mean free path.

203. Gas exerts pressure on the walls of container

because the molecules-

(1) Are loosing their Kinetic energy

(2) Are getting stuck to the walls

(3) Are transferring their momentum to walls

(4) Are accelerated towards walls.

204. For a diatomic gas, change in internal energy for

unit change in temperature at constant pressure

and volume is U1 and U2 respectively then U1 : U2

is :

(1) 5 : 3 (2) 7 : 5

(3) 1 : 1 (4) 5 : 7

205. The specific heat of an ideal gas depends on

temperature is -

(1) 1T

(2) T

(3) T

(4) Does not depends on temperature

206. The specific heat of a gas :

(1) Has only two value Cp and Cv

(2) Has a unique value at a given temperature

(3) Can have any value between 0 and ¥

(4) Depends upon the mass of the gas

207. 22 g of CO2 at 270C is mixed with 16 g of O2 at

370C. The temperature of the mixture is :–

(At room temperature, degrees of freedom of

CO2=7 and degrees of freedom of O2 = 5)

(1) 31.160C (2) 270C

(3) 370C (4) 300C

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

90 E

Pre-Medical : Physics ALLEN208. At 270C temperature, the kinetic energy of an ideal

gas is E1. If the temperature is increased to 3270C,

then kinetic energy would be

(1) 2 E1 (2) 1

1E

2

(3) 12 E (4) 1

1E

2

209. Oxygen and hydrogen gases are at temperature

T. Then K.E of molecules of oxygen gas is equal to

how many times of average K.E. of molecules of

hydrogen gas :-

(1) 16 times (2) 8 times

(3) Equal (4) 1/16 times

210. The average energy of the molecules of a

monoatomic gas at temperature T is :-

( K = Boltzmann constant )

(1) 12

kT (2) kT

(3) 32

kT (4) 52

kT

211. A diatomic molecule has

(1) 1 degree of freedom




212. Two moles of monoatomic gas are mixed with

1 mole of a diatomic gas. Then g for the mixture is:

(1) 1.4 (2) 1.55 (3) 1.62 (4) 1.67

ZEROTH AND FIRST LAW OF THERMODYNAMICS,

HEAT, WORK AND INTERNAL ENERGY

213. The first law of thermodynamics is based on :–

(1) Law of conservation of energy

(2) Law of conservation of mechanical energy

(3) Law of conservation of gravitational P.E.


214. In a process, 500 calories of heat is given to a sys-

tem and at the same time 100 joules of work is

done on the system. The increase in the internal

energy of the system is :-

(1) 40 calories (2) 1993 joules

(3) 2193 joules (4) 82 calories

215. In a thermodynamic process pressure of a fixed mass

of a gas is changed in such a manner that the gas

releases 20 joules of heat and 8 joules of work was

done on the gas. If the initial internal energy of the

gas was 30 joules, then the final internal energy

will be:-

(1) 2 J (2) 42 J (3) 18 J (4) 58 J

216. When a system is taken from state ‘a’ to state

‘b’ along the path ‘acb’, it is found that a quantity

of heat Q = 200 J is absorbed by the system and

a work W = 80J is done by it. Along the path

‘adb’, Q = 144J. The work done along the path

‘adb’ is

pc

a

b

dV

(1) 6J (2) 12 J (3) 18 J (4) 24 J

217. In the above question, if the work done on the

system along the curved path ‘ba’ is 52J, heat

absorbed is

(1) –140 J (2) –172 J

(3) 140 J (4) 172 J

218. In question No. 217, if Ua = 40J, value of Ubwill

be

(1) – 50 J (2) 100 J

(3) –120 J (4) 160 J

219. In question No. 217, if Ud = 88 J, heat absorbed

for the path ‘db’ is

(1) –72 J (2) 72 J

(3) 144 J (4) –144 J

220. 1 kg of a gas does 20 kJ of work and receives

16 kJ of heat when it is expanded between two

states. A second kind of expansion can be found

between the same initial and final state which

requires a heat input of 9 kJ. The work done by the

gas in the second expansion is :

(1) 32 kJ (2) 5 kJ

(3) –4 kJ (4) 13 kJ

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

91E

Pre-Medical : PhysicsALLEN221. As shown in the figure the amount of heat absorbed

along the path ABC is 90J and the amount of work

done by the system is 30 J. If the amount of work

done along the path ADC is 20 J then amount of

heat absorbed will be :-

B

A

C

D

P

V

(1) 80 J (2) 90 J (3) 110 J (4) 120 J

222. In a cyclic process shown on the P – V diagram, the

magnitude of the work done is :

P1

P2

O V1 V2

V

P

(1) 2

2 1P P2-æ öpç ÷

è ø(2)

2

2 1V V2-æ öpç ÷

è ø

(3) 4p

(P2 – P1) (V2 – V1) (4) p (P2V2 – P1V1)

223. The work by an ideal monoatomic gas along the

cyclic path LMNOL is

(1) PV

(2) 2PV

(3) 3 PV

(4) 4 PV

224. For a gas Cv = 4.96 cal/mole K, the increase in

internal energy of 2 mole gas in heating from

340 K to 342 K will be :-

(1) 27.80 cal (2) 19.84 cal

(3) 13.90 cal (4) 9.92 cal

225. When a system changes from one to another state

the value of work done :-

(1) Depends on the force acting on the system

(2) Depends on the nature of material present in a

system

(3) Does not depend on the path

(4) Depends on the path

226. A system is taken along the paths A and B as shown.

If the amounts of heat given in these processes are

QA and QB and change in internal energy are DUA

and DUB respectively then :-

(1) QA = QB; DUA < DUB A

B

i fP

V

(2) QA ³ QB; DUA = DUB

(3) QA < QB; DUA > DUB

(4) QA > QB; DUA = DUB

227. If the heat of 110 J is added to a gaseous system

and change in internal energy is 40 J, then the

amount of external work done is :

(1) 180 J (2) 70 J (3) 110 J (4) 30 J

228. If amount of heat supplied is Q, work done is W

and change in internal energy is mCV dT, then

relation among them is. (Cv = gram specific heat)

(1) mCV dT = Q + W

(2) Q = W + mCV dT

(3) Q + mCV dT = W

(4) None of these.

229. The work done by a gas taken through the closed

process ABCA is

A

BC

5P0

P0

2V0 5V0

V

P

(1) 6P0V0 (2) 4P0V0 (3) P0V0 (3) zero

230. In the given figure, the initial and final states of a

gas are shown by points i and f. The internal energy

of the gas at i is 10J. For the path iaf : dQ = 50 J

and dW = 20 J. For the path ibf, if dQ= 36 J the

value of dW will be equal to

Pa f

i b

® V

(1) 4J (2) 12J (3) 18J (4) 6J

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

92 E

Pre-Medical : Physics ALLENCYCLIC, ISOCHORIC, ISOBARIC, ISOTHERMAL,ADIABATIC AND POLYTROPIC PROCESS

231. In the diagrams (i) to (iv) of variation of volume with

changing pressure is shown. A gas is taken along

the path ABCDA. The change in internal energy of

the gas will be:-

(i)

A B

D C

V

P

(ii)

V

P

A B

D C

(iii)

A B

D CV

P

(iv)

V

P

A B

D C

(1) Positive in all cases (i) to (iv)

(2) Positive in cases (i), (ii) and (iii) but zero in case (iv)

(3) Negative in cases (i), (ii) and (iii) but zero in case (iv)

(4) Zero in all the four cases

232. The temperature of 5 moles of a gas which was

held at constant volume was changed from 1000C

to 1200C. The change in internal energy was found

to be 80 joules. The total heat capacity of the gas

at constant volume will be equal to:-

(1) 8 J/K (2) 0.8 J/K

(3) 4.0 J/K (4) 0.4 J/K

233. Monoatomic, diatomic and triatomic gases whose

initial volume and pressure are same, are

compressed till their volume become half the initial

volume.

(1) If the compression is adiabatic then monoatomic

gas will have maximum final pressure.

(2) If the compression is adiabatic then triatomic

gas will have maximum final pressure.

(3) If the compression is adiabatic then their final

pressure will be same.

(4) If the compression is isothermal then their final

pressure will be different.

234. Which of the following graphs correctly represents

the variation of b = –(dV/dP)/V with P for an ideal

gas at constant temperature?

(1) (2)

(3) (4)

235. The adiabatic Bulk modulus of a diatomic gas at

atmospheric pressure is

(1) 0 Nm–2

(2) 1 Nm–2

(3) 1.4 ×104 Nm–2

(4) 1.4 ´ 105 Nm–2

236. A given quantity of an ideal gas is at pressure P and

absolute temperature T. The isothermal bulk modulus

of the gas is :

(1) 2P/3 (2) P (3) 3P/2 (4) 2P

237. P-V plots for two gases during adiabatic processes

are shown in the figure. Plots 1 and 2 should

correspond respectively to

(1) He and O2 (2) O2 and He

(3) He and Ar (4) O2 and N2

238. An ideal gas undergoes the process 1 ® 2 as shown

in the figure, the heat supplied and work done in

the process is DQ and DW respectively. The ratio

DQ : DW is

(1) g : g – 1 (2) g(3) g – 1 (4) g – 1 : g

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

93E

Pre-Medical : PhysicsALLEN239. For an adiabatic expansion of a perfect gas, the

value of DP/P is equal to:-

(1) – g DV/V (2) –DV/V

(3) –g DV/V (4) –g2 DV/V

240. An ideal gas at 270C is compressed adiabatically to

8/27 of its original volume. If g = 5/3, then the rise

in temperature is:-

(1) 450 K (2) 375 K (3) 675 K (4) 405 K

241. For monoatomic gas the relation between pressure

of a gas and temperature T is given by P µ TC .

Then value of C will be : (For adiabatic process)

(1) 53

(2) 25

(3) 35

(4) 52

242. A gas for which g = 5/3 is heated at constant

pressure. The percentage of total heat given that

will be used for external work is :

(1) 40% (2) 30% (3) 60% (4) 20%

243. In which of the figure no heat exchange between

the gas and the surroundings will take place, if the

gas is taken along curve:

(curves are isothermal and adiabatic)

A

B D

C

Pre

ssur

e

Volume

(1) A (2) B (3) C (4) D

244. In the following figures, four curves A, B, C, D are

shown the curves are :–

A

B

Volume

Pre

ssur

e

Volume

Pre

ssur

e C

D

(1) Isothermal for A and B while adiabatic for

C and D

(2) Isothermal for A and C while adiabatic for

B and D

(3) Isothermal for A and D

(4) Adiabatic for A and C while isothermal for

B and D

245. Equal volumes of a perfect gas are compressed to

half of their initial volumes. The first is brought about

by isothermal process and the second by adiabatic

process then :

(1) Both temperature and pressure will increase in

the isothermal process.

(2) In the isothermal process, the temperature will

decrease and pressure will increases

(3) Both temperature and pressure will increase in

adiabatic process

(4) In the adiabatic process, the temperature will

decrease and pressure will increase

246. A vessel contains an ideal monoatomic gas which

expands at constant pressure, when heat Q is given

to it. Then the work done in expansion is :

(1) Q (2) 35

Q (3) 25

Q (4) 23

Q

247. One mole of an ideal gas at temperature T1 expends

according to the law 2

PV

= a (constant). The work

done by the gas till temperature of gas becomes T2

is :

(1) 12

R(T2 – T1) (2) 13

R(T2 – T1)

(3) 14

R(T2 – T1) (4) 15

R(T2 – T1)

248. When an ideal diatomic gas is heated at constant

pressure, the fraction of the heat energy supplied

which increases the internal energy of the gas is -

(1) 2/5 (2) 3/5 (3) 3/7 (4) 5/7

249. Pressure-temperature relationship for an ideal gas

undergoing adiabatic change is (g = Cp/Cv)

(1) gPT = constant

(2) - +g1PT = constant

(3) g- g1PT T = constant

(4) -g g1P T = constant

250. The value of internal energy in an adiabatic

process :-

(1) Remains unchanged

(2) Only increases

(3) Only diminishes

(4) May diminish and may also increase

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

94 E

Pre-Medical : Physics ALLEN251. One mole of an ideal monoatomic gas is heated at

a constant pressure of one atmosphere from 0°C

to 100°C. Then the change in the internal energy is

(1) 20.80 × 102 J

(2) 12.48 × 102 J

(3) 832 × 102 J

(4) 6.25 × 102 J

252. The specific heat of a gas at constant pressure is

more than that of the same gas at constant volume

because :-

(1) Work is done in the expansion of gas at constant

pressure

(2) Work is done in the expansion of the gas at

contant volume

(3) The molecular attraction incrcase under constant

pressure

(4) The vibration of molecules increases under

constant pressure

253. When a gas is adiabatically compressed then it's

temperature increase because :–

(1) Work done is zero

(2) Internal energy is increased

(3) Heat is supplied to it

(4) No change in pressure

254. Air is filled in a tube of the wheel of a car at 27°C

and 2 atm pressure if the tube is suddenly bursts,

the final temperature of air will be :–

(g = 1.5, 21/3 = 1.251)

(1) – 33° C (2) 0° C

(3) 21.6° C (4) 240° C

255. Specific heat of a gas undergoing adiabatic change

is :

(1) Zero (2) Infinite

(3) Positive (4) Negative

256. A quantity of air (g = 1.4) at 27°C is compressed

suddenly, the temperature of the air system will :

(1) Fall

(2) Rise

(3) Remain unchanged

(4) First rise and then fall

257. A gas speciman in one vessel is expended

isothermally to double its volume and a simillar

specimen in the second vessel is expanded

adiabatically the same extent, then :

(1) In the second vessel, both pressure and work

done are more

(2) In the second vessel, pressure in more, but the

work done isless.

(3) In the first vessel, both pressure & work done

are more.

(4) In the first vessel, pressure is more, but work

done is less

258. If an ideal gas is compressed during isothermal

process then :-

(1) No work is done against gas

(2) heat is rejected by gas

(3) It's internal energy will increase

(4) Pressure does not change

259. Graphs between P–V diagram for isothermal and

adiabatic processes are drawn the relation between

their slopes will be :–

(1) Slope of adiabatic curve=g (slope of isothermal

curve)

(2) Slope of isothermal curve = g (slope of adiabatic

curve)

(3) Slope of isothermal curve = slope of adiabatic curve

(4) Slope of adiabatic curve = g2 (slope of isother-

mal curve)

260. An ideal gas is taken round the cycle ABCA. In the

cycle the amount of work done involved is :–

P1

4P1

C B

3V1V1

AP

V

(1) 12 P1V1 (2) 6 P1V1

(3) 3 P1V1 (4) P1V1

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

95E

Pre-Medical : PhysicsALLEN261. One mole ideal gas is compressed adiabatically at

27°C. Its temperature becomes 102°C. The work

done in this process will be :- (g = 1.5)

(1) –625 J (2) 625 J

(3) 1245 J (4) –1245 J

262. In an isometric change :

(1) Q dUd = (2) W dUd =

(3) Q W dUd + d = (4) None of these

263. The volume of a gas expands by 0.25 m3 at a

constant pressure of 103 N/m2 . The work done is

equal to

(1) 2.5 erg (2) 250 J

(3) 250 W (4) 250 N

264. The process in which the heat given to a system is

completely transformed into work is for ideal gas :–

(1) Isobaric process (2) Isometric process

(3) Isothermal process (4) Adiabatic process

265. The volume of a poly-atomic gas 43

æ ög =ç ÷è ø

compressed adiabatically to th

18

of the original

volume. If the original pressure of the gas is P0 the

new pressure will be :

(1) 8 P0 (2) 16 P0

(3) 6 P0 (4) 2 P0

266. During isothermal, isobaric and adiabatic processes,

work done for same change in volume will be

maximum for :-

(1) Isothermal (2) Isobaric

(3) Adiabatic (4) None of the above

267. In an adiabatic process the quantity which remains

constant is :-

(1) Temperature

(2) Pressure

(3) Total heat content of the system

(4) Volume

268. Graph of isometric process is :-

(1) P

T

(2) P

T

(3) P

T

(4) P

T269. An ideal gas undergoes an adiabatic change in

volume (V) with pressure (P). Then :-

(1) PgV = constant (2) PVg = constant

(3) (PV)g = constant (4) PV = constant

270. 300 calories of heat is supplied to raise the

temperature of 50 gm of air from 20°C to 30°C

without any change in its volume. Change in internal

energy per gram of air is

(1) zero (2) 0.6 calories

(3) 1.2 calories (4) 6.0 calories

271. A gas is expanded from volume V1 to volume V2

in three processes, shown in the figure. If UA, UB,

UC and UD represent the internal energies of the

gas in state A,B, C and D respectively, the which

of the following is not correct

P

® V

Isobaric B

Isothermal

Adiabatic

D

V1 V2

C

A

(1) UB – UA > 0 (2) UC – UA = 0

(3) UD – UA <0 (4) UB = UC = UD

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

96 E

Pre-Medical : Physics ALLEN272. A closed container is fully insulated from outside.

One half of it is filled with an ideal gas X separated

by a plate P from the other half Y which contains a

vacuum as shown in figure. When P is removed, X

moves into Y. Which of the following statements is

correct?

Xgas

Yvacuum

p

(1) No work is done by X

(2) X decreases in temperature

(3) X increases in internal energy

(4) X doubles in pressure

SECOND LAW OF THE THERMODYNMAMICS,

HEAT ENGINES AND REFRIGERATORS

273. According to the second law of thermodynamics :

(1) heat energy cannot be completely converted

to work

(2) work cannot be completely converted to heat

energy

(3) for all cyclic processes we have dQ/T < 0

(4) the reason all heat engine efficiencies are less

than 100% is friction, which is unavoidable

274. "Heat cannot flow by itself from a body at lower

temperature to a body at higher temperature" is a

statement or consequence of :

(1) second law of thermodynamics

(2) conservation of momentum

(3) conservation of mass

(4) first law of thermodynamics

275. A Carnot engine takes 3 × 106 cal of heat from

reservoir at 627°C and gives it to a sink at 27°C.

Then work done by the engine is

(1) 4.2 × 106 J (2) 8.4 × 106 J

(3) 16.8 × 106 J (4) zero

276. A reversible refrigerator operates between a low

temperature reservoir at TC and a high temperature

reservoir at TH. Its coefficient of performance is

given by :

(1) (TH – TC)/TC (2) TC/(TH – TC)

(3) (TH – TC)/TH (4) TH/(TH – TC)

277. In the given graph the isothermal curves are :-

(1) AB and CD (2) AB and AD

(3) AD and BC (4) BC and CD

278. In the above question, the curve for which the heat

is absorbed from the surroundings is :–

(1) AB (2) BC (3) CD (4) DA

279. A carnot engine shows efficiency of 40% on taking

energy at 500 K. To increase the efficiency to 50%,

at what temperature it should take energy ?

(1) 400 K (2) 700 K (3) 600 K (4) 800 K

280. A Carnot engine, whose efficiency is 40%, takes in

heat from a source maintained at a temperature of

500 K. It is desired to have an engine of efficiency

60%. Then, the intake temperature for the same

exhaust (sink) temperature must be :

(1) 750 K

(2) 600 K

(3) Efficiency of Carnot engine cannot be made

larger than 50%

(4) 1200 K

281. If the system takes 100 cal. heat, and releases

80 cal to sink, if source temperature is 127°C find

the sink temperature :–

(1) 47° C (2) 127° C (3) 67° C (4) 107° C

282. A carnot engine working between 300 K and

600 K has work out put of 800 J per cycle. The

amount of heat energy supplied to the engine from

source per cycle will be :

(1) 800 J (2) 1600 J

(3) 1200 J (4) 900 J

283. The efficiency of carnot engine is 50% and

temperature of sink is 500K. If temperature of

source is kept constant and its efficiency raised to

60%, then the required temperature of the sink

will be :-

(1) 100 K (2) 600 K (3) 400 K (4) 500 K

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

97E

Pre-Medical : PhysicsALLEN284. An ideal gas heat engine operates in carnot cycle

between 227°C and 127°C. It absorbs 6 × 104 cal

of heat at higher temperature. Then amount of heat

converted to work is :

(1) 2.4 × 104 cal (2) 6 × 104 cal

(3) 1.2 × 104 cal (4) 4.8 × 104 cal

EXERCISE-I (Conceptual Questions) ANSWER KEY

285. A refrigerator works between temperature –10°C

and 27°C, the coefficient of performance is :

(1) 7.1 (2) 1

(3) 8.1 (4) 15.47

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 1 1 1 2 4 3 3 1 2 2 2 1 3 4 1Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. 3 1 1 1 1 4 2 4 1 1 3 3 4 3 1Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 4 4 3 1 2 3 2 2 2 3 1 1 1 2 1Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. 4 4 2 1 3 2 2 2 4 2 3 1 1 2 2Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75Ans. 4 3 3 1 3 3 3 1 3 1 4 3 4 1 2Que. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90Ans. 2 3 2 4 3 3 3 3 4 2 4 2 4 4 2Que. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105Ans. 1 1 1 2 3 3 1 3 2 2 3 1 4 1 3Que. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120Ans. 4 4 3 2 4 3 1 1 2 3 3 2 1 2 2Que. 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135Ans. 4 1 4 4 4 4 4 3 1 3 3 3 4 3 2Que. 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150Ans. 3 3 2 3 2 4 4 3 4 3 4 3 4 4 2Que. 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165Ans. 1 3 3 3 2 1 2 1 1 3 2 2 2 2 1Que. 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180Ans. 1 1 4 3 1 4 1 4 2 4 3 4 1 2 4Que. 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195Ans. 4 1 4 3 2 4 4 3 4 2 1 3 1 3 4Que. 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210Ans. 3 3 3 4 3 2 4 3 3 4 3 1 1 3 3Que. 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225Ans. 3 2 1 3 3 4 2 4 2 4 1 3 1 2 4Que. 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240Ans. 4 2 2 1 4 4 3 1 1 4 2 2 1 3 2Que. 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255Ans. 4 1 2 2 3 3 2 4 4 4 2 1 2 1 1Que. 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270Ans. 2 3 2 1 3 4 1 2 3 2 2 3 3 2 4Que. 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285Ans. 4 1 1 1 2 2 1 1 3 1 1 2 3 3 1

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

98 E


AIPMT 20061. A black body emits radiation of maximum intensity

at 5000 Å when its temperature is 1227° C. If its

temperature is increased by 1000° C then the

maximum intensity of emitted radiation will be at:

(1) 2754.8 Å (2) 3000 Å

(3) 3500 Å (4) 4000 Å

2. The translational kinetic energy of molecules of one

mole of a monoatomic gas is U=3NkT/2. The value

of molar specific heat of gas under constant pressure

will be :

(1) 32

R (2) 52

R

(3) 72

R (4) 92

R

3. The molar specific heat at constant pressure of an

ideal gas is (7/2)R. The ratio of specific heat at

constant pressure to that at constant volume is :

(1) 75

(2) 87

(3) 57

(4) 97

AIIMS 2006

4. Three objects coloured black, gray and white can

withstand hostile conditions upto 2800°C. These

objects are thrown into a furnace where each of

them attains a temperature of 2000°C. Which

object will gloss brightest :-

(1) The white object

(2) The black object

(3) All glow with equal brightness

(4) Gray object

5. Two balloons are filled, one with pure He gas and

the other by air, respectively. If the pressure and

temperature of these balloons are same then the

number of molecules per unit volume is :-

(1) more in the He filled balloon

(2) same in both balloons

(3) more in air filled balloon

(4) in the ratio of 1 : 4

AIPMT 2007

6. An ideal monoatomic gas is taken round the cycle

ABCDA as shown in following P - V diagram. The

work done during the cycle is :

A

D

B

C

P

VO

3P,V 3P,3V

P,V P,3V

(1) PV (2) 2 PV (3) 4 PV (4) Zero

7. The WQ

æ öç ÷è ø

of a carnot-engine is 16

, now the

temperature of sink is reduced by 62°C, then this

ratio becomes twice, therefore the init ial

temperature of the sink and source are respectively:-

(1) 33°C, 67°C (2) 37°C, 99°C

(3) 67°C, 33°C (4) 97 K, 37 K

AIPMT 2008

8. A new scale of temperature (which is linear) called

the W scale, the freezing and boiling points of water

are 39°W and 239°W respectively. What will be the

temperature on the new scale, corresponding to a

temperature of 39°C on the Celsius scale ?

(1) 200° W (2) 139° W

(3) 78° W (4) 117°W

9. At 10°C the value of the density of a fixed mass of

an ideal gas divided by its pressure is x. At 110°C

this ratio is :-

(1)10

x110

(2) 283

x383

(3) x (4) 383

x283

10. If Q, E and W denote the heat added, change in

internal energy and the work done respectively in

a closed cycle process, then :-

(1) E = 0 (2) Q = 0

(3) W=0 (4) Q=W=0

EXERCISE-II (Previous Year Questions) AIPMT/NEET & AIIMS (2006-2018)

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

99E

Pre-Medical : PhysicsALLENAIPMT 2009

11. The two ends of a rod of length L and a uniform

cross sect ional area A are kept at two

temperatures T1 and T2(T1 > T2). The rate of heat

transfer dQdt

, through the rod in a steady state

is given by :–

(1) 1 2kA(T T )dQdt L

-= (2) 1 2kL(T T )dQ

dt A-

=

(3) 1 2k(T T )dQdt LA

-= (4) 1 2

dQkLA(T T )

dt= -

12. A black body, at a temperature of 227°C radiates

heat at a rate of 7 cal cm–2 s–1. At a temperature

of 727°C, the rate of heat radiated in the same

units will be :–

(1) 80 (2) 60 (3) 50 (4) 112

13. In thermodynamic processes which of the following

statement is not true :-

(1) In an adiabatic process PVg = constant

(2) In an adiabatic process the system is insulated

from the surroundings

(3) In an isochoric process pressure remains constant

(4) In an Isothermal process the temperature remains

constant

14. The change in internal energy in a system that has

absorbed 2 Kcals of heat and 500 J of work done

is :-

(1) 7900J (2) 8900J

(3) 6400J (4) 5400J

AIPMT (Pre) 2010

15. A cylindrical metallic rod in thermal contact with two

reservoirs of heat at its two ends conducts an amount

of heat Q in time t. The metallic rod is melted and

the material is formed into a rod of half the radius

of the original rod. What is the amount of heat

conducted by the new rod, when placed in thermal

contact with the two reservoirs in time t ?

(1) Q2

(2) Q4

(3) Q16

(4) 2Q

16. Total radiant energy per unit area, per unit timenormal to the direction of incidence, received at adistance R from the centre of a star of radius r,whose outer surface radiates as a black body at atemperature T Kelvin is given by :-

(1) ps 2 4

2

4 r T

R(2)

s 2 4

2

r T

R

(3) s

p

2 4

2

r T

4 r(4)

s 4 4

4

r T

r(Where s is Stefan's Constant)

17. If DU and DW represent the increase in internalenergy and work done by the system respectivelyin a thermodynamic process, which of the followingis true ?(1) DU = –DW, in a isothermal process(2) DU = –DW, in a adiabatic process(3) DU = DW, in a isothermal process

(4) DU = DW, in a adiabatic process

AIPMT (Mains) 2010

18. If cp and cv denote the specific heats (per unit mass)

of an ideal gas of molecular weight M, then :-

(1) cp – cv = R (2) cp – cv = RM

(3) cp – cv = MR (4) cp – cv = 2

R

Mwhere R is the molar gas constant

19. A monoatomic gas at pressure P1 and volume V1

is compressed adiabatically to 1/8th its original

volume. What is the final pressure of the gas :-

(1) P1 (2) 16P1 (3) 32 P1 (4) 64 P1

AIPMT (Pre) 2011

20. During an isothermal expansion, a confined idealgas does +150 J of work against its surroundings.This implies that :-(1) 150 J of heat has been removed from the gas(2) 300 J of heat has been added to the gas(3) No heat is transferred because the process is

isothermal(4) 150 J of heat has been added to the gas

21. A mass of diatomic gas (g = 1.4) at a pressure of2 atmospheres is compressed adiabatically so thatits temperature rises from 27°C to 927°C. Thepressure of the gas in the final state is :-(1) 8 atm (2) 28 atm

(3) 68.7 atm (4) 256 atm

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

100 E

Pre-Medical : Physics ALLEN22. When 1kg of ice at 0°C melts to water at 0°C, the

resulting change in its entropy, taking latent heat

of ice to be 80 cal/g, is -

(1) 273 cal/K (2) 8 × 104 cal/K

(3) 80 cal/K (4) 293 cal/K

AIPMT (Pre) 2012

23. If the radius of a star is R and it acts as a black body,

what would be the temperature of the star, in which

the rate of energy production is Q ?

(1) (4pR2Q/s)1/4 (2) (Q/4pR2s)1/4

(3) Q/4pR2s (4) (Q/4pR2s)–1/2

(s stands for Stefan's constant.)

24. One mole of an ideal gas goes from an initial state

A to final state B via two processes. It firstly

undergoes isothermal expansion from volume V to

3V and then its volume is reduced from 3V to V at

constant pressure. The correct P-V diagram

representing the two processes is :-

(1)

P

VV 3V

A

B

(2)

P

A

VV 3V

B

(3)

P A

VV 3V

B

(4)

P

A

VV 3V

B

25. A thermodynamic system is taken through the cycle

ABCD as shown in figure. Heat rejected by the gas

during the cycle is :-

Pre

ssur

e

VolumeV 3V

A B

D C2P

P

(1) 12

PV (2) PV (3) 2PV (4) 4PV

26. Liquid oxygen at 50 K is heated to 300 K at constant

pressure of 1 atm. The rate of heating is constant.

Which one of the following graphs represents the

variation of temperature with

time ?

(1)

Temp

Time

(2)

Temp

Time

(3)

Temp

Time

(4)

Temp

Time

AIPMT (Mains) 2012

27. A slab of stone of area 0.36 m2 and thickness

0.1 m is exposed on the lower surface to steam at

100°C. A block of ice at 0°C rests on the upper

surface of the slab. In one hour 4.8 kg of ice is

melted. The thermal conductivity of slab is :

(Given latent heat of fusion of ice 3.36 × 105 J kg–1)

(1) 2.05 J/m/s/°C (2) 1.02 J/m/s/°C

(3) 1.24 J/m/s/°C (4) 1.29 J/m/s/°C

28. An ideal gas goes from state A to state B via three

different processes as indicated in the P-V diagram

3 B

1AP

V

2

If Q1 , Q2, Q3 indicate the heat absorbed by the gas

along the three processes and DU1, DU2, DU3

indicate the change in internal energy along the

three processes respectively, then :-

(1) Q1 = Q2 = Q3 and DU1 > DU2 > DU3

(2) Q3 > Q2 > Q1 and DU1 > DU2 > DU3

(3) Q1 > Q2 > Q3 and DU1 = DU2 = DU3

(4) Q3 > Q2 > Q1 and DU1 = DU2 = DU3

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

101E

Pre-Medical : PhysicsALLENAIIMS 2012

29. Two conductors having thickness d1 and d2,

thermal conductivity k1 and k2 are placed one

above the another. Find the equivalent thermal

conductance :-

(1) 1 2 1 2 2 1

1 2

(d d )(k d k d )2(k k )

+ ++

T1

T2

d1

d2

(2) 1 2 1 2 2 1

1 2

(d d )(k d k d )2(k k )

- ++

(3) 1 2 1 2

1 2 2 1

k k (d d )d k d k

++

(4) None of these

30. Degree of freedom for polyatomic gas :-

(1) ³ 3 (2) ³ 4 (3) ³ 5 (4) ³ 7

31. Conversion of water to steam is accompained by

which process :-

(1) Adiabatic (2) Isothermal

(3) Isochoric (4) Cyclic

32. Calculate the work done for B ® A, :-

A

C B

0 1 5 P(N/m )2

2

4

V(L)

(1) 6 × 10–3 J (2) 12 × 10–3 J

(3) 3 × 10–3 J (4) 4 × 10–3 J

33. What is the slope for an isothermal process in PV

indicator diagram :-

(1) PV

(2) P

–V

(3) Zero (4) ¥

NEET-UG 2013

34. A piece of iron is heated in a flame. It first becomes

dull red then becomes reddish yellow and finally turns

to white hot. The correct explanation for the above

observation is possible by using :-

(1) Newton's Law of cooling

(2) Stefan's Law

(3) Wein's displacement Law

(4) Kirchoff's Law

35. In the given (V – T) diagram, what is the relation

between pressure P1 and P2 ?

q2

q1

P2

P1

T

V

(1) Cannot be predicted

(2) P2 = P1

(3) P2 > P1

(4) P2 < P1

36. The amount of heat energy required to raise the

temperature of 1 g of Helium at constant volume,

from T1 K to T2 K is :-

(1) 34

Na kB æ öç ÷è ø

2

1

TT

(2) 38

Na kB (T2 – T1)

(3) 32

Na kB (T2 – T1)

(4) 34

Na kB (T2 – T1)

37. The molar specific heats of an ideal gas at constant

pressure and volume are denoted by CP and CV

respectively. If g = P

V

CC

and R is the universal gas

constant, then CV is equal to :

(1) gR (2) + g- g

11

(3) ( )g -R

1 (4) ( )g -1

R

38. During an adiabatic process, the pressure of a gas

is found to be proportional to the cube of its

temperature. The ratio of p

v

C

C for the gas is :-

(1) 32

(2) 43

(3) 2 (4) 53

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

102 E

Pre-Medical : Physics ALLEN39. A gas is taken through the cycle A®B®C®A, as

shown, What is the net work done by the gas ?

2 4 6 8

123

54

6

0

7B

CA

P(10 Pa)5

V(10 m )–3 3

(1) –2000 J (2) 2000 J

(3) 1000 J (4) Zero

AIIMS 2013

40. In a closed cylinder there are 60 g Ne and

64 g O2. If pressure of mixture of gases in cylinder

is 30 bar then partial pressure of O2 in this cylinder

will be :-

(1) 30 bar (2) 20 bar

(3) 15 bar (4) 12 bar

41. A gas mixture contain one mole He gas and one

mole O2 gas. Find the ratio of specific heat at

constant pressure to that at constant volume of the

gaseous mixture :-

(1) 2 (2) 1.5

(3) 2.5 (4) 4

42. One mole of oxygen of volume 1 litre at 4 atm

pressure to attain 1 atm pressure by result of

isothermal expansion. Find work done by the gas.

(1) ~155 J (2) ~206 J

(3) ~355 J (4) ~562 J

43. Which of the following is a state function :-

(1) Work done in cyclic process

(2) Work done in isothermal process

(3) Heat at constant pressure

(4) Heat at constant volume

44. A refrigerator transfer 180 joule of energy in one

second from temperature –3°C to 27°C. Calculate

the average power consumed, assuming no energy

losses in the process.

(1) 18 W (2) 54 W

(3) 20 W (4) 120 W

AIPMT 2014

45. Steam at 100°C is passed into 20 g of water at

10°C. When water acquires a temperature of 80°C,

the mass of water present will be :

[Take specific heat of water = 1 cal g–1 °C–1 and

latent heat of steam = 540 cal g–1]

(1) 24 g (2) 31.5 g

(3) 42.5 g (4) 22.5 g

46. Certain quantity of water cools from 70°C to 60°C

in the first 5 minutes and to 54°C in the next 5

minutes. The temperature of the surroundings is:-

(1) 45°C (2) 20°C (3) 42°C (4) 10°C

47. The mean free path of molecules of a gas, (radius 'r')

is inversely proportional to :-

(1) r3 (2) r2 (3) r (4) r

48. A monoatomic gas at a pressure P, having a volume

V expands isothermally to volume 2V and then

adibatically to volume 16V. The final pressure of the

gas is : (take g = 53

)

(1) 64P (2) 32P (3) P64

(4) 16P

49. A thermodynamic system undergoes cyclic process

ABCDA as shown in fig. The work done by the

system in the cycle is :-

A D

C B

V0 2V0

P0

2P0

3P0

P

V

(1) P0V0 (2) 2P0V0

(3) 0 0P V2

(4) Zero

AIIMS 2014

50. Two stars A and B of surface area Sa and Sb &

temperature Ta and Tb glow red and blue

respectively. Choose the correct option.

(1) Ta > Tb (2) Ta < Tb

(3) TaSa = TbSb (4) TaSb = TbSa

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

103E

Pre-Medical : PhysicsALLEN51. A gas is equilibrium at T kelvin. If mass of one

molecule is m and its component of velocity in x

direction is vx. Then mean of its vx2 is

(1) 3kTm

(2) 2kTm

(3) kTm

(4) zero

52. A diatomic gas undergoes adiabatic compression

and its volume reduces to half of initial volume then

its final pressure would be if initial pressure of gas

is P.

(1) 21.4P (2) P/2

(3) 2P (4) 3.07 P

AIPMT 2015

53. One mole of an ideal diatomic gas undergoes a

transition from A to B along a path AB as shown

in the figure,

A

B

5

2

4 6

V(m )3

P(kPa)

The change in internal energy of the gas during the

transition is :

(1) –20 kJ (2) 20 J (3) –12 kJ (4) 20 kJ

54. On observing light from three different stars P, Q

and R, it was found that intensity of violet color is

maximum in the spectrum of P, the intensity of green

colour is maximum in the spectrum of R and the

intensity of red colour is maximum in the spectrum

of Q. If TP, TQ and TR are the respective absolute

temperatures of P, Q and R, then it can be

concluded from the above observation that :

(1) TP > TR > TQ (2) TP < TR < TQ

(3) TP < TQ < TR (4) TP > TQ > TR

55. A Carnot engine, having an efficiency of h = 1

10as

heat engine, is used as a refrigerator. If the work

done on the system is 10 J, the amount of energy

absorbed from the reservoir at lower temperature

is :-

(1) 99 J (2) 90 J (3) 1 J (4) 100 J

56. The ratio of the specific heats = gP

V

CC

in terms of

degrees of freedom (n) is given by :

(1) æ ö+ç ÷è ø

n1

3(2)

æ ö+ç ÷è ø

21

n

(3) æ ö+ç ÷è ø

n1

2(4)

æ ö+ç ÷è ø

11

n

57. Figure below shows two paths that may be taken by

a gas to go from a state A to a state C.

V

2×10 m–3 3 4×10 m–3 3

2×10 Pa4

6×10 Pa4

P B C

A

In process AB, 400 J of heat is added to the system

and in process BC, 100 J of heat is added to the

system. The heat absorbed by the system in the

process AC will be :

(1) 500 J (2) 460 J

(3) 300 J (4) 380 J

58. The two ends of a metal rod are matainted at

temperatures 100°C and 110°C. The rate of heat

flow in the rod is found to be 4.0 J/s. If the ends

are maintained at temperatures 200°C and 210°C,

the rate of heat flow will be :

(1) 16.8 J/s (2) 8.0 J/s

(3) 4.0 J/s (4) 44.0 J/s

Re-AIPMT 2015

59. Two vessels separately contain two ideal gases A and

B at the same temperature, the pressure of A being

twice that of B. Under such conditions, the density

of A is found to be 1.5 times the density of B. The

ratio of molecular weight of A and B is:

(1) 12

(2) 23

(3) 34

(4) 2

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

104 E

Pre-Medical : Physics ALLEN60. 4.0 g of a gas occupies 22.4 litres at NTP. The

specific heat capacity of the gas at constant volumeis 5.0 JK–1 mol–1 . If the speed of sound in this gasat NTP is 952 ms–1, then the heat capacity atconstant pressure is

(Take gas constant R = 8.3 JK–1 mol–1)

(1) 8.5 JK–1 mol–1 (2) 8.0 JK–1 mol–1

(3) 7.5 JK–1 mol–1 (4) 7.0 JK–1 mol–1

61. The coefficient of performance of a refrigerator is5. If the temperature inside freezer is –20°C, thetemperature of the surroundings to which it rejectsheat is :

(1) 21°C (2) 31°C (3) 41°C (4) 11°C

62. An ideal gas is compressed to half its initial volumeby means of several processes. Which of the processresults in the maximum work done on the gas ?

(1) Isothermal (2) Adiabatic

(3) Isobaric (4) Isochoric

63. The value of coefficient of volume expansion ofglycerin is 5 × 10–4 K–1. The fractional change inthe density of glycerin for a rise of 40°C in itstemperature, is :-

(1) 0.010 (2) 0.015 (3) 0.020 (4) 0.025

AIIMS 2015

64. One of the most efficient engines ever developedoperated between 2100 K and 700 K. Its actualefficiency is 40%. In what percentage of its actualefficiency of its maximum possible efficiency is(approx).

(1) 40% (2) 60% (3) 75% (4) 90%

65. In Maxwell's velocity distrbution curve area under thegraph

(1) Increases when temperature is increased

(2) deccreases when temperature is increased(3) remains same at all temperature

(4) depends on the pressure of the gas

66. In a gas at 1 atm pressure and 27°C a moleculehas diameter of 5Å find mean free path ofmolecules:-

(1) 3.8 × 10–8 m (2) 2 × 10–8 m

(3) 8.3 × 10–9 m (4) 3.2 × 10–6 m

67. A body cool from 90°C to 70°C in 10 minutes iftemperature of surrounding is 20°C find the timetaken by body to cool from 60°C to 30°C. AssumingNewton's law of cooling is valid.

(1) 10 min (2) 24 min

(3) 36 min (4) 8 min

NEET-I 2016

68. A refrigerator works between 4°C and 30°C. It isrequired to remove 600 calories of heat everysecond in order to keep the temperature of therefrigerated space constant. The power required is:

(Take 1 cal = 4.2 Joules)

(1) 2.365 W (2) 23.65 W

(3) 236.5 W (4) 2365 W

69. A black body is at a temperature of 5760 K. Theenergy of radiation emitted by the body atwavelength 250 nm is U1, at wavelength 500 nmis U2 and that at 1000 nm is U3. Wien's constant,b = 2.88 × 106 nmK. Which of the following iscorrect?

(1) U1 = 0 (2) U3 = 0

(3) U1 > U2 (4) U2 > U1

70. Coefficient of linear expansion of brass and steelrods are a1 and a2. Lengths of brass and steel rodsare l1 and l2 respectively. If (l2 – l1) is maintainedsame at all temperatures, which one of the followingrelations holds good ?

(1) a1l2 = a2l1

(2) a1l22 = a2l1

2

(3) a12l2 = a2

2l1

(4) a1l1 = a2l2

71. The molecules o f a given mass of a gashave r.m.s. velocity of 200 m/s at 27°C and1.0 × 105 N/m2 pressure. When the temperatureand pressure of the gas are respectively, 127°C and0.05 × 105 N/m2, the r.m.s. velocity of its moleculesin m/s is :

(1) 100 2 (2) 400

3

(3) 100 2

3(4)

1003

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

105E

Pre-Medical : PhysicsALLEN72. A gas is compressed isothermally to half its initial

volume. The same gas is compressed separately

through an adiabatic process until its volume is again

reduced to half. Then :-

(1) Compressing the gas isothermally will require

more work to be done.

(2) Compressing the gas through adiabatic process

will require more work to be done.

(3) Compressing the gas i sothermally or

adiabatically will require the same amount of

work.

(4) Which of the case (whether compression through

isothermal or through adiabatic process) requires

more work will depend upon the atomicity of

the gas.

73. A piece of ice falls from a height h so that it melts

completely. Only one-quarter of the heat produced

is absorbed by the ice and all energy of ice gets

converted into heat during its fall. The value

of h is :

[Latent heat of ice is 3.4 × 105 J/kg and

g = 10 N/kg]

(1) 34 km

(2) 544 km

(3) 136 km

(4) 68 km

NEET-II 2016

74. Two identical bodies are made of a material for

which the heat capacity increases with temperature.

One of these is at 100 °C, while the other one is

at 0°C. If the two bodies are brought into contact,

then, assuming no heat loss, the final common

temperature is :-

(1) less than 50 °C but greater than 0 °C

(2) 0 °C

(3) 50 °C

(4) more than 50 °C

75. A body cools from a temperature 3T to 2T in

10 minutes. The room temperature is T. Assume

that Newton's law of cooling is applicable. The

temperature of the body at the end of next

10 minutes will be :-

(1)4

T3

(2) T

(3) 7

T4

(4) 3

T2

76. One mole of an ideal monatomic gas undergoes a

process described by the equation PV3 = constant.

The heat capacity of the gas during this process

is

(1) 2 R (2) R

(3) 3

R2

(4) 5

R2

77. The temperature inside a refrigerator is t2°C and the

room temperature is t1°C. The amount of heat

delivered to the room for each joule of electrical

energy consumed ideally will be :-

(1)+-

2

1 2

t 273t t (2)

++1 2

1

t tt 273

(3) -1

1 2

tt t (4)

+-

1

1 2

t 273t t

78. A given sample of an ideal gas occupies a volume

V at a pressure P and absolute temperature T. The

mass of each molecule of the gas is m. Which of the

following gives the density of the gas ?

(1) P/(kTV) (2) mkT

(3) P/(kT) (4) Pm/(kT)

AIIMS 2016

79. In adiabatic process, volume of monoatomic gas

increases by 6% then find percentage change in

temperature.

(1) –2% (2) –4%

(3) 2% (4) 4%

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

106 E

Pre-Medical : Physics ALLEN80. 5 moles of a gas expand from V to 2V at

400 K. Find work done by gas.

(1) 11.50 kJ (2) 23.4 kJ

(3) 63.7 kJ (4) 5.6 kJ

81. A tungsten body of diameter 2.3 cm is at 2000°C.

It radiates 30% of the energy radiated by a black

body of same radius and temperature. Find radius

of black body which will radiate energy at same rate

at the same temperature ?

(1) 2.32 cm (2) 1.49 cm

(3) 0.629 cm (4) 0.123 cm

82.

1

0.5

P(atm)

10 50V(litre)

Find out work done by gas in cyclic process

(1) 100 J (2) 200 J

(3) 600 J (4) 1000 J

83. When 250 KJ heat is given to the system then it does

work of 400 KJ then calculate change in internal

energy of system ?

(1) 150 KJ (2) –150 KJ

(3) 200 KJ (4) –200 KJ

NEET(UG)-2017

84. Two rods A and B of different materials are welded

together as shown in figure. Their thermal

conductivities are K1 and K2. The thermal

conductivity of the composite rod will be :-

T1 T2

d

(1) +1 23(K K )2

(2) K1 + K2

(3) 2 (K1 + K2) (4) +1 2K K2

85. Thermodynamic processes are indicated in the

following diagram :

f

f

f700k

500k300k

f

P

I

IV

IIIII

V

i

Match the following

Column-1 Column-2P. Process I a. Adiabatic

Q. Process II b. Isobaric

R. Process III c. Isochoric

S. Process IV d. Isothermal

(1) P ® c, Q ® a, R ® d, S ® b

(2) P ® c, Q ® d, R ® b, S ® a

(3) P ® d, Q ® b, R ® a, S ® c

(4) P ® a, Q ® c, R ® d, S ® b

86. A spherical black body with a radius of 12 cmradiates 450 watt power at 500 K. If the radius werehalved and the temperature doubled, the powerradiated in watt would be :-

(1) 450 (2) 1000

(3) 1800 (4) 225

87. A carnot engine having an efficiency of 1

10 as heat

engine, is used as a refrigerator. If the work done on

the system is 10 J, the amount of energy absorbed

from the reservoir at lower temperature is :-

(1) 90 J (2) 99 J

(3) 100 J (4) 1 J

88. A gas mixture consists of 2 moles of O2 and 4 moles

of Ar at temperature T. Neglecting all vibrational

modes, the total internal energy of the system is:-

(1) 15 RT (2) 9 RT (3) 11 RT (4) 4 RT

AIIMS 2017

89. For an isolated system

(1) Q = 0 , W = 0 (2) Q ¹ 0 , W = 0

(3) Q = 0 , W ¹ 0 (4) Q ¹ 0 , W ¹ 0

90. Ideal gas at pressure P undergoes free expansionfrom its initial volume of 1.01 Litre to 10.1 Litre.The final pressure of the gas is?

(1) P(10)g (2) P

10g

(3) 10P (4) P

10

91. Argon gas is at initial temperature 20°C. If gas isexpanded adiabatically 8 times to its initial volumethen find final temperature of Argon gas?

(1) –100°C (2) –50°C

(3) –200°C (4) 0

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

107E

Pre-Medical : PhysicsALLEN92. At constant pressure how much fraction of heat

supplied to gas is converted into mechanical work?

(1)1g -

g (2) 1

gg -

(3) g –1 (4) 1

gg +

NEET(UG)-201893. A sample of 0.1 g of water at 100oC and normal

pressure (1.013 ´ 105 Nm–2) requires 54 cal of heat

energy to convert to steam at 100oC. If the volume

of the steam produced is 167.1 cc, the change in

internal energy of the sample, is :-

(1) 104.3 J (2) 208.7 J

(3) 42.2 J (4) 84.5 J

94. The power radiated by a black body is P and it

radiates maximum energy at wavelength l0. If the

temperature of the black body is now changed so

that it radiates maximum energy at wavelength

l0

34

, the power radiated by it becomes nP. The

value of n is :-

(1) 34

(2) 43

(3) 25681

(4) 81

256

95. The volume (V) of a monatomic gas varies with its

temperature (T), as shown in the graph. The ratio

of work done by the gas, to the heat absorbed by

it, when it undergoes a change from state A to state

B, is :-

AB

TO

V

(1) 25

(2) 23

(3) 13

(4) 27

96. The efficiency of an ideal heat engine working

between the freezing point and boiling point of

water, is :-

(1) 26.8% (2) 20% (3) 6.25% (4) 12.5%

97. At what temperature will the rms speed of oxygen

molecules become just sufficient for escaping from

the Earth's atmosphere ?

(Given :

Mass of oxygen molecule (m) = 2.76 × 10–26 kg

Boltzmann's constant kB = 1.38 × 10–23 J K–1) :-

(1) 2.508 × 104 K (2) 8.360 × 104 K

(3) 5.016 × 104 K (4) 1.254 × 104 K

AIIMS 2018

98. Which of the following properties of athermodynamic system are extensive ?(1) V and T (2) P and T(3) V and E (4) E and Cp

99. 1 cm3 of water at 100°C and 1 atm pressure isheated to convert it to 1521 cm3 vapour. Findincrease in internal energy if latent heat ofvaporization of water = 540 cal/g.(1) 3.26 kJ (2) 2.12 kJ(3) 4.18 kJ (4) 1.05 kJ

100. Which of the following is intensive property of a

thermodynamic system ?

(1) P and T (2) V and T

(3) V and P (4) V and E

101. Work done in expanding one mole of an ideal gas

isothermally from 2L to 4L is equal to the work done

in expanding three moles of an ideal gas from 2L to

x L at same temperature. Find x.

(1) 81/3 (2) 42/3 (3) 2 (4) 16

102. Consider the given temperature entropy diagram.

Which of the following processes represent increase

in volume ?

A B

D C

T

SO

(1) DA and AB (2) AB and BC

(3) BC and CD (4) CD and DA

103. Nitrogen gas is contained in a cylinder at

8 atm pressure and 300 K. The density of nitrogen

in the container is nearly :-

(1) 7.9 kg/m3 (2) 9 kg/m3

(3) 1.3 kg/m3 (4) 12 kg/m3

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

108 E

Pre-Medical : Physics ALLEN104. An ideal gas undergoes a cyclic process as shown in

diagram. The net work done by the gas in the cycle

is :-

4 12 V (litre)

5

2

0

P (atm)

(1) 12 litre-atm (2) 24 J

(3) 24 litre-atm (4) –24 J

105. A thermodynamic system does 60 J of useful work.

If its internal energy decreases by 50J, then how

much heat has been supplied to the system ?

(1) 10 J (2) 110 J (3) –10 J (4) –60 J

106. In an adiabatic expansion of 0.5 mole of an ideal

diatomic gas, temperature changes from 32°C to

20°C. The change in internal energy of the gas is

(1) 124.65 cal. (2) –124.65 J

(3) 72.32 J (4) 248.3 J

107. A refrigerator having coefficient of performance 10

absorbs 500 J per cycle. The amount of heat

rejected to the surrounding per cycle is :-

(1) 50 J (2) 550 J

(3) 450 J (4) 500 J

108. Two gases A and B have same temperature. Density

of A is twice the density of B but molar mass of B is

twice as that of A. The ratio of pressures of gases A

and B is :-

(1) 1 : 1 (2) 2 : 1

(3) 4 : 1 (4) 1 : 4

109. The change in entropy of 1 kg of nitrogen gas in an

isobaric process when its temperature changes from

400 K to 800 K is :-

(1) 720 J/K (2) 420 J/K

(3) 309 J/K (4) 720 cal/°C

ANSWER KEYQue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Ans. 2 2 1 2 2 3 2 4 2 1 1 4 3 1 3Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Ans. 2 2 2 3 4 4 4 2 2 3 3 3 3 3 3Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

Ans. 2 1 2 3 4 2 3 1 3 4 2 4 4 3 4Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

Ans. 1 2 3 4 2 3 1 1 1 2 2 2 3 3 2

Que. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Ans. 2 2 3 2 3 1 3 3 4 4 2 2 3 4 4Que. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

Ans. 2 4 4 2 1 3 4 2 4 1 3 1 3 1 4Que. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105Ans. 3 1 2 3 1 1 2 3 2 1 2 2 2 3 1Que. 106 107 108 109Ans. 2 2 3 1

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

109E

Pre-Medical : PhysicsALLENEXERCISE-III (Analytical Questions) Check Your Understanding

1. Two rods one of aluminium of length l1 having

coefficient of linear expansion aa, and other steel

of length l2 having coefficient of linear expansion

as are joined end to end. The expansion in both the

rods is same on variation of temperature. Then the

value of 1

1 2+l

l l is

(1) s

a s

aa + a (2)

s

a s

aa - a

(3) a s

s

a + aa (4) None of these

2. An ideal gas is expanding such that PT2 = constant.

The coefficient of volume expansion of the gas is :

(1) 1T

(2) 2T

(3) 3T

(4) 4T

3. Steam at 100°C is added slowly to 1400 g of water

at 16°C until the temperature of water is raised to

80°C. The mass of steam required to do this is

(LV = 540 cal/g) :

(1) 160 g (2) 125 g

(3) 250 g (4) 320 g

4. 50 g of ice at 0° C is mixed with 50 g of water at

100° C. The final temperature of mixture is :–

(1) 0° C

(2) Between 0°C to 20°C

(3) 20°C

(4) Above 20°C

5. In an insulated vessel, 0.05 kg steam at 373 K and

0.45 kg of ice at 253 K are mixed. Then, find the

final temperature of the mixture.

Given, Lfusion = 80 cal/g = 336 J/g, Lvaporization =

540 cal/g = 2268 J/g, Sice = 2100 J/kg K = 0.5

cal/gK and Swater = 4200 J/kg K = 1 cal /gK

(1) 273 K (2) 373 K

(3) 100 K (4) 0 K

6. Spheres P and Q are uniformly constructed from

the same material which is a good conductor of heat

and the radius of Q is thrice of radius of P. The rate

of fall of temperature of P is x times that of Q when

both are at the same surface temperature. The value

of x is :

(1) 1/4 (2) 1/3 (3) 3 (4) 4

7. A sphere, a cube and a thin circular plate all made

of same substance and all have same mass. These

are heated to 200°C and then placed in a room,

then the :

(1) Temperature of sphere drops to room

temperature at last.

(2) Temperature of cube drops to room temperature

at last

(3) Temperature of thin circular plate drops to room

temperature at last

(4) Temperature of all the three drops to room

temperature at the same time

8. The power radiated by a black body is P and it

radiates maximum energy around the wavelength

l0. If the temperature of the black body is now

changed so that it radiates maximum energy around

wavelength 3/4l0, the power radiated by it will

increase by a factor of

(1) 4/3 (2) 16/9

(3) 64/27 (4) 256/81

9. A slab consists of two parallel layers of copper and

brass of the same thickness and having thermal

conductivities in the ratio 1 : 4. If the free face of

brass is at 100°C and that of copper at 0°C, the

temperature of interface is -

(1) 80°C (2) 20°C (3) 60°C (4) 40°C

10. Out of the metal balls of same diameter one is solid

and other is hollow. Both are heated to the same

temperature at 300°C and then allowed to cool in

the same surroundings then rate of loss of heat will

be:

(1) More for hollow sphere

(2) More for solid sphere

(3) Same for both


Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

110 E

Pre-Medical : Physics ALLEN11. For a black body at temperature 727°C, its radiating

power is 60 watt and temperature of surrounding

is 227°C. If temperature of black body is changed

to 1227°C then its radiating power will be :-

(1) 304 W (2) 320 W

(3) 240 W (4) 120 W

12. A body takes T minutes to cool from 62°C to 61°C

when the surrounding temperature is 30°C. The

time taken by the body to cool from 46°C to 45.5°C

is :–

(1) Greater than T minutes

(2) Equal to T minutes

(3) Less than T minutes

(4) Equal to T/2 minutes

13. Three discs A, B, and C having radii 2 m, 4 m and

6 m respectively are coated with carbon black on

their outer surfaces. The wavelengths corresponding

to maximum intensity are 300 nm, 400 nm and

500 nm respectively. The power radiated by them

are QA, QB and QC respectively then

(1) QA is maximum (2) QB is maximum

(3) QC is maximum (4) QA = QB = QC

14. The spectral emissive power El for a body at

temperature T1 is plotted against the wavelength

and area under the curve is found to be A. At a

different temperature T2, the area is found to be

9A. Then l1/l2 =

(1) 3 (2) 1/3 (3) 1 3 (4) 3

15. Consider a gas with density r and c as the root

mean square velocity of its molecules contained in

a volume. If the system moves as whole with velocity

v, then the pressure exerted by the gas is

(1) 13

r ( )c 2 (2) 13

r( c + v)2

(3) 13

r( c – v)2 (4) 13

r(c–2 – v)2

16. Air is filled at 600C in a vessel of open mouth. The

vessel is heated to a temperature T so that

1/4th part of air escapes. The value of T is :

(1) 800C (2) 4440C

(3) 3330C (4) 1710C

17. Root mean square velocity for a certain di-atomic gas at

room temperature 270C is found to be 1930 m/sec.

The gas is -

(1) H2 (2) O2 (3) F2 (4) Cl2

18. The equation of state of a gas is given by

2aTP

V

æ ö+ç ÷

è øVC = (RT + b), where a, b, c and R are

constants. This isotherms can be represented by

P = AVm – BVn, where A and B depend only on

temperature and

(1) m = – c and n = – 1

(2) m = c and n = 1

(3) m = – c and n = 1

(4) m = c and n = – 1

19. An ideal gas expands in such a way that

PV2 = constant throughout the process.

(1) The graph of the process of T-V diagram is a

parabola.

(2) The graph of the process of T-V diagram is a

straight line.

(3) Such an expansion is possible only with heating.

(4) Such an expansion is possible only with cooling.

20. Two identical glass bulbs are interconnected by a

thin glass tube at 0ºC. A gas is filled at N.T.P. in

these bulb is placed in ice and another bulb is placed

in hot bath, then the pressure of the gas becomes

1.5 times. The temperature of hot bath will be :–

Ice Hot water

(1) 100°C (2) 182°C

(3) 256°C (4) 546°C

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

111E

Pre-Medical : PhysicsALLEN21. A gas has volume V and pressure P. The total

translational kinetic energy of all the molecules of

the gas is:–

(1) 32

PV only if the gas is monoatomic.

(2) 32

PV only if the gas is diatomic.

(3) >32

PV if the gas is diatomic.

(4) 32

PV in all cases.

22. N molecules of an ideal gas at temperature T1 and

presseure P1 are contained in a closed box. If the

molecules in the box gets doubled, Keeping total

kinetic energy same. If new pressure is p2 and

temperature is T2, Then :

(1) P2 = P, T2 = T1

(2) P2 = P1, T2 = T1 / 2

(3) P2 = 2 P1, T2 = T1

(4) P2 = 2P1, T2 = T1 / 2

23. The following are the P–V diagrams for cyclic

processes for a gas. In which of these processes

heat is released by the gas ?

(1)

V

P (2) V

P

(3)

V

P (4) V

P

24. A gas mixture contain 1 g H2 and 1 g He if

temperature of gas mixture is increased from 0°C

to 100° C at isobaric process. Then find given heat

of gas mixture

[gHe = 5/3, gH2

= 7/5, R = 2 cal/mol-K]

(1) 124 cal (2) 327 cal

(3) 218 cal (4) 475 cal

25. An ideal gas expands isothermally from a volume

V1 to V2 and then compressed to original volume

V1 adiabatically. Initial pressure is P1 and final

pressure is P3. The total work done is W. Then

(1) P3 > P1, W > 0

(2) P3 < P1, W < 0

(3) P3 > P1, W < 0

(4) P3 = P1, W = 0

26. An ideal gas is taken through the cycle

A ® B ® C ® A, as shown in the figure. If the net

heat supplied to the gas in the cycle is 5J, the work

done by the gas in the process C ® A is

(1) –5J (2) –10 J (3) –15 J (4) –20 J

27. One mole of a monatomic ideal gas is taken through

a cycle ABCDA as shown in the P-V diagram.

Column-II gives the characteristics involved in the

cycle. Match them with each of the processes given

in Column-I.

DC

B A

P

3P

1P

0 1V 3V 9V V

Column-I Column-II(A) Process A ® B (p) Internal energy decreases

(B) Process B ® C (q) Internal energy increases

(C) Process C ® D (r) Heat is lost

(D) Process D ® A (s) Heat is gained

(t) Work is done on the gas

(1) A ® p,r,t ; B ® p,r ; C ® q,s ; D ® r,t

(2) A ® r,t ; B ® q,r ; C ® q,s ; D ® r,s

(3) A ® p,r,t; B ® p,q,r ; C ® s ; D ® r,t

(4) A ® p,r,t ; B ® p,r ; C ® q ; D ® r,s

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

112 E


EXERCISE-III (Analytical Questions) ANSWER KEY

Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 1 3 1 2 1 3 1 4 1 3 2 2 2 4 1Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. 4 1 1 4 4 4 2 4 4 3 1 1 1 3 3Que. 31 32 33

Ans. 3 2 2

28. Three processes form a thermodynamic cycle as

shown on P-V diagram for an ideal gas. Process

1 ® 2 takes place at constant temperature (300K).

Process 2 ® 3 takes place at constant volume.

During this process 40J of heat leaves the system.

Process 3 ® 1 is adiabatic and temperature T3 is

275K. Work done by the gas during the process

3 ®1 is

(1) -40J (2) -20J (3) +40J (4) +20J

29. The molar specific heat under constant pressure of

oxygen is CP = 7.03 cal/mol K. The quantity of

heat required to raise the temperature from 100C

to 200C of 5 moles of oxygen under constant

volume will approximately be :-

(1) 25 cal (2) 50 cal

(3) 250 cal (4) 500 cal

30. If the ratio of specific heat of a gas at constant

pressure to that at constant volume is g, the change

in internal energy of a mass of gas, when the volume

changes from V to 2 V at constant pressure P, is :-

(1) R/( g - 1) (2) PV

(3) PV/(g - 1) (4) gPV/(g - 1)

31. The molar heat capacity in a process of a diatomic

gas, if it does a work of Q4

when a heat of Q is

supplied to it, is :-

(1) 2

R5

(2) 5

R2

(3) 10

R3

(4) 6

R7

32. The above P-V diagram represents the

thermodynamic cycle of an engine, operating with

an ideal monoatomic gas. The amount of heat,

extracted from the source in a single cycle is :

2P0

P

P0

V0 2V0V

A

B C

D

(1) P0V0 (2) 0 013

P V2

æ öç ÷è ø

(3) 0 011

P V2

æ öç ÷è ø

(4) 4P0V0

33. In the above question efficiency of cycle ABCDA is

nearly :

(1) 12.5% (2) 15.4%

(3) 9.1% (4) 10.5%

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

113E

Pre-Medical : PhysicsALLENEXERCISE-IV (Assertion & Reason) Target AIIMS

These questions consist of two statements each, printed as Assertion and Reason. While answering these Questions you are required to choose any one of the following four responses.

(A) If both Assertion & Reason areTrue & theReason is a correct explanation of the Assertion.

(B) If both Assertion & Reason areTrue but Reason is not a correct explanation of the Assertion.

(C) If Assertion is True but the Reason is False.

(D) If both Assertion & Reason are false.

Directions for Assertion & Reason questions

1. Assertion : Liquids usually expand more than solids.

Reason : The intermolecular forces in liquids are

weaker than in solids.

(1) A (2) B (3) C (4) D

2. Assertion : Rubber contract on heating.

Reason : In rubber as temperature increases, the

amplitude of transverse vibrations increases more

than the amplitude of longitudinal vibrations.

(1) A (2) B (3) C (4) D

3. Assertion : A temperature change which increases

the length of a steel rod by 1% will increase its volume

by 3%.

Reason : The coefficient of volume expansion is

nearly three times the coeff icient of linear

expansion.

(1) A (2) B (3) C (4) D

4. Assertion : The melting point of ice decreases with

increases of pressure.

Reason : Ice contracts on melting. [AIIMS 2004]

(1) A (2) B (3) C (4) D

5. Assertion : Water kept in an open vessel will

quickly evaporate on the surface of the moon.

Reason : There is lack of atmospheric pressure at

surface of moon.

(1) A (2) B (3) C (4) D

6. Assertion : In a pressure cooker the water is

brought to boil. The cooker is then removed from

the stove. Now, on removing the lid of the pressure

cooker, the water starts boiling again. [AIIMS 2004]

Reason : When we add small impurities in water,

the boiling point of water will decrease.

(1) A (2) B (3) C (4) D

7. Assertion : In pressure-temperature phase diagram

of water, the slope of the melting curve in found to

be negative. [AIIMS 2005]

Reason : Ice contracts on melting of water.

(1) A (2) B (3) C (4) D

8. Assertion : A solid material is supplied heat at

constant rate. The temperature of the material is

changing with the heat input as shown in Figure.

Temp.

O

A B

C DE

Heat input

If CD = 2AB then latent heat of vaporisation of

substance is double that of fusion.

Reason : Latent heat of fusion µ AB and Latent

heat of vaporisation µ CD

(1) A (2) B (3) C (4) D

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

114 E

Pre-Medical : Physics ALLEN9. Assertion : It is hotter over the top of a fire than at

the same distance on the sides. [AIIMS 2003]

Reason : Air surrounding the fire transfer more

heat upwards.

(1) A (2) B (3) C (4) D

10. Assertion : Coolant coils are fitted at the top of a

refrigerator, for formation of convection current.

Reason : Air becomes denser on cooling.

(1) A (2) B (3) C (4) D

11. Assertion : Perspiration from human body helps

in cooling the body. [AIIMS 2006]

Reason : A thin layer of water on the skin enhances

its emissivity.

(1) A (2) B (3) C (4) D

12. Assertion : A sphere, a cube and a thin circular

plate made of same material and of same mass are

initially heated to 200°C, the plate will cool at fastest

rate.

Reason : Rate of cooling =eAmss 4 4

0(T T )- µ surface

area and surface area is maximum for plate.

(1) A (2) B (3) C (4) D

13. Assertion : Blue star is at high temperature than

red star.

Reason : According to Wien's displacement law

lm = bT '

where symbols have their usual meanings.

(1) A (2) B (3) C (4) D

14. Assertion : A body that is a good radiator is also a

good absorber of radiation at a given wavelength.

Reason : According to Kirchhoff's law the

absorptivity of a body is equal to its emissivity at a

given wavelength. [AIIMS 2005]

(1) A (2) B (3) C (4) D

15. Assertion : For higher temperature, the peak

emission wavelength of a black body shifts to lower

wavelengths. [AIIMS 2005]

Reason : Peak emission wavelength of a black body

is proportional to the fifth power of temperature.

(1) A (2) B (3) C (4) D

16. Assertion : While measuring the thermal

conductivity of liquid experimentally, the upper layer

is kept hot and the lower layer is kept cold.

Reason : This avoids heating of liquid by convection.

[AIIMS 2007]

(1) A (2) B (3) C (4) D

17. Assertion : The amount of radiation from sun's

surface varies as the fourth power of its absolute

temperature.

Reason : The sun is black body. [AIIMS 1999]

(1) A (2) B (3) C (4) D

18. Assertion : If a gas container in motion is suddenly

stopped, the temperature of the gas rises.

Reason : In given process, the kinetic energy of

ordered mechanical motion is converted into the

kinetic energy of random motion of gas molecules.

(1) A (2) B (3) C (4) D

19. Assertion : A material will have only one specific

heat, if and only if its coefficient of thermal

expansion is equal to zero.

Reason : An ideal gas has two specific heats

(CV and CP) only.

(1) A (2) B (3) C (4) D

20. Assertion : The value of DQ is always zero in

adiabatic process. [AIIMS 2010]

Reason : Adiabatic process is always a cyclic

process.

(1) A (2) B (3) C (4) D

21. Assertion : First law of thermodynamics allows

many processes which actually don't happen.

Reason : First law of thermodynamics must not be

violated for any process to happen.

(1) A (2) B (3) C (4) D

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

115E

Pre-Medical : PhysicsALLEN22. Assertion : Thermodynamics processes in nature

are irreversible. [AIIMS 2004]

Reason : Dissipative effects cannot be eliminated.

(1) A (2) B (3) C (4) D

23. Assertion : Reversible system are difficult to find

in real world. [AIIMS 2005]

Reason : Most processes are dissipative in nature.

(1) A (2) B (3) C (4) D

24. Assertion : Air quickly leaking out of a ballon

becomes cooler. [AIIMS 2005]

Reason : The leaking air undergoes adiabatic

expansion.

(1) A (2) B (3) C (4) D

25. Assertion : When a glass of hot milk is placed in a

room and allowed to cool, its entropy decreases.

Reason : Allowing hot object to cool does not violate

the second law of thermodynamics.

[AIIMS 2006]

(1) A (2) B (3) C (4) D

26. Assertion : The Carnot cycle is useful in

understanding the performance of heat engines.

Reason : The Carnot cycle provides a way of

determinig the maximum possible efficiency

achievable with reservoirs of given temperatures.

[AIIMS 2006]

(1) A (2) B (3) C (4) D

27. Assertion : In an isolated system, the entropy

increases.

Reason : The processes in an isolated system are

adiabatic. [AIIMS 2006]

(1) A (2) B (3) C (4) D

28. Assertion : Two isothermal curves intersect each

other at a certain point. [AIIMS 2008]

Reason : The isothermal change are done slowly,

so the isothermal curves have very little slope.

(1) A (2) B (3) C (4) D

29. Assertion : It is not possible for a system, unaided

by an external agency to transfer heat from a body

at lower temperature to another at a higher

temperature. [AIIMS 1994]

Reason : It is not possible to violate the second law

of thermodynamics.

(1) A (2) B (3) C (4) D

30. Assertion : It is impossible for a ship to use the

internal energy of sea water to operate its engine.

Reason : A refrigerator is a heat engine working in

the reverse direction. [AIIMS 2009]

(1) A (2) B (3) C (4) D

31. Assertion : When a bottle of cold carbonated drink

is opened, a slight fog forms around the opening.

Reason : Adiabatic expansion of the gas causes

lowering of temperature which start condensation

of water vapours. [AIIMS 2003]

(1) A (2) B (3) C (4) D

32. Assertion : In isothermal process heat energy

supplied to system is converted completely into

work.

Reason : According to first law of thermodynamics

Q = W + DU.

(1) A (2) B (3) C (4) D

33. Assertion : The curve A and B in figure, show

P-V graphs for an isothermal and an adiabatic

process for an ideal gas. The isothermal process is

represented by B.

A

B

P

V

Reason : On P-V graph, modulus of slope of the

adiabatic curve is greater then the modulus of the

slope of the isothermal curve.

(1) A (2) B (3) C (4) D

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

116 E

Pre-Medical : Physics ALLEN34. Assertion : During rapid pumping of air in tyres,

air inside the tyre is hotter than atmospheric air.

Reason : Adiabatic process occurs at very high

rate.

(1) A (2) B (3) C (4) D

35. Assertion : The root mean square and most

probable speeds of the molecules in a gas are the

same. [AIIMS 2006]

Reason : The maxwell's distribution for the speed

of molecules in a gas is symmetrical.

(1) A (2) B (3) C (4) D

36. Assertion : The ratio P

V

CC

for a monoatomic gas is

more then for a diatomic gas. [AIIMS 1998]

Reason : The molecules of a monoatomic gas have

more degree of freedom then those of a diatomic

gas.

(1) A (2) B (3) C (4) D

37. Assertion : Air pressure in a car tyre increases

during driving. [AIIMS 2007]

Reason : Absolute zero temperature is not zero

energy temperaure.

(1) A (2) B (3) C (4) D

38. Assertion : The ratio P

V

CC

is more for helium gas

than for hydrogen gas.

Reason : Atomic mass of helium is more than that

of hydrogen. [AIIMS 1996]

(1) A (2) B (3) C (4) D

39. Assertion : For an ideal gas CP > CV.

Reason : Work is done in expansion of the gas at

constant pressure. [AIIMS 2011]

(1) A (2) B (3) C (4) D

40. Assertion : For an ideal gas, at constant

temperature, the product of the pressure and

volume is constant. [AIIMS 1998]

Reason : The mean square velocity of gas molecules

is inversely proportional to its mass.

(1) A (2) B (3) C (4) D

41. Assertion : Water cannot be boiled inside a satellite

by convection.

Reason : In weightlessness conditions, natural

movement of heated fluid is not possible.

(1) A (2) B (3) C (4) D

42. Assertion :- The value of specific heat at constant

pressure is more than that at constant volume.

Reason :- The energy required to raise

temperature by a unit at constant pressure is greater

because some amount of heat is used in doing work.

[AIIMS 2015]

(1) A (2) B (3) C (4) D

43. Assertion :- A good absorber is a good emitter.

Reason :- For a substance, absorbtance is equal

to emissivity. [AIIMS 2015]

(1) A (2) B (3) C (4) D

44. Assertion :- Sum of two path functions is always

a path function.

Reason :- Every state function can be produced by

path functions. [AIIMS 2015]

(1) A (2) B (3) C (4) D

45. Assertion :- Specific heat of metals approaches

zero as temperature approaches zero kelvin.

Reason :- As tempeature approaches zero kelvin,

effectiveness of degree of freedom decreaes.

[AIIMS 2015]

(1) A (2) B (3) C (4) D

46. Assertion :- When we apply brakes to car its

entropy doesn't increase. [AIIMS 2016]

Reason :- It is a reversible process.

(1) A (2) B (3) C (4) D

47. Assertion :- A fluid in vacuum expands such that

it changes to gaseous state. Its temperature

decreases. [AIIMS 2016]

Reason :- It uses its internal energy as heat of

vaporization.

(1) A (2) B (3) C (4) D

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

117E

Pre-Medical : PhysicsALLEN48. Assertion :- H2 is found more as compared to N2

in earth's upper atmosphere. [AIIMS 2016]

Reason :- N2 is heavier than H2.

(1) A (2) B (3) C (4) D

49. Assertion :- PV value of gas can increase onincreasing the pressure. [AIIMS 2016]

Reason :- Pressure always increases oncompression.

(1) A (2) B (3) C (4) D

50. Assertion :- Heat is not a state theromodynamicfunction. [AIIMS 2016]

Reason :- Heat given in a system depends onprocess.

(1) A (2) B (3) C (4) D

51. Assertion :- In an adibatic process, temperatureof system decrease when work is done by thesystem. [AIIMS 2016]

Reason :- A part of internal energy is utilized indoing work.

(1) A (2) B (3) C (4) D

52. Assertion :- For an isolated system, mechanicalenergy is not conserved. [AIIMS 2017]

Reason :- For an isolated system internal energyis constant.

(1) A (2) B (3) C (4) D

53. Assertion :- Adiabatic free expansion is not areversible process. [AIIMS 2017]

Reason :- In adiabatic expansion internal energydoes not change.

(1) A (2) B (3) C (4) D

54. Assertion :- Generally, molecules which havemaximum velocity are lesser in number.

Reason :- The molecules of the gas followMaxwell's speed distribution. [AIIMS 2017]

(1) A (2) B (3) C (4) D

55. Assertion :- Adiabatic free expansion of an idealgas is irreversible. [AIIMS 2017]

Reason :- PV = constant for reversible adiabaticexpansion.

(1) A (2) B (3) C (4) D

56. Assertion :- Thermal energy and heat energy are

same. [AIIMS 2018]

Reason :- Both energies depend on randomnessof molecules.

(1) A (2) B (3) C (4) D

57. Assertion :- Temperature is constant in a Quasi-

static isothermal process. [AIIMS 2018]

Reason :- No heat is exchanged with surroundingin a quasi-static isothermal process.

(1) A (2) B (3) C (4) D

58. Assertion :- For an isolated system, internal energy

remains constant. [AIIMS 2018]

Reason :- Heat is exchanged through walls of

isolated system .

(1) A (2) B (3) C (4) D

59. Assertion :- Between states (P1, V1, T1) and

(P2, V2, T2), work done in two step process is more

than work done in one step process. [AIIMS 2018]

Reason :- Work done is a path function.

(1) A (2) B (3) C (4) D

60. Assertion :- In process of throttling, temperature

neither increases nor decreases. [AIIMS 2018]

Reason :- Throttling is an isothermal process.

(1) A (2) B (3) C (4) D

61. Assertion :- In an adiabatic process, work done

by an ideal gas equals change in its inernal energy.

Reason :- In an adiabatic process, temperature of

system and surrounding remains same.

[AIIMS 2018]

(1) A (2) B (3) C (4) D

62. Assertion :- On increasing temperature, internal

energy of monoatomic gas increases.

Reason :- Thermal energy increases only random

motion of gas molecules. [AIIMS 2018]

(1) A (2) B (3) C (4) D

63. Assertion :- In a thermoflask, inner surface is shiny.

Reason :- Shiny surface is a poor radiator of heat.

(1) A (2) B (3) C (4) D

64. Assertion :- In free expansion of an ideal gas

internal energy is constant. [AIIMS 2018]

Reason :- No heat is added and no work is done in

free expansion.

(1) A (2) B (3) C (4) D

Z:\N

ODE

02\B

0AI-B

0\TA

RGET

\PHY

\EN

G\M

OD

ULE_

03\0

1-TH

ERM

AL P

HYSI

CS\0

2- E

XERC

ISE.

P65

118 E

Pre-Medical : Physics ALLEN65. Assertion :- Pressure in a real gas is less than that

in an ideal gas. [AIIMS 2018]

Reason :- Force of attraction is non-zero in a real

gas.

(1) A (2) B (3) C (4) D

66. Assertion :- Food is cooked with difficulty at

mountain peak. [AIIMS 2018]

Reason :- Water boils at high temperature

there.

(1) A (2) B (3) C (4) D

EXERCISE-IV (Assertion & Reason) ANSWER KEYQue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Ans. 1 1 1 1 1 3 1 1 1 1 3 1 1 1 3Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30Ans. 1 1 1 3 3 2 1 1 1 2 1 4 4 1 2Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45Ans. 1 1 1 1 4 3 2 2 1 2 1 1 1 4 1Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans. 4 1 1 3 1 1 4 3 1 3 4 3 3 4 3Que. 61 62 63 64 65 66 67 68Ans. 3 3 1 1 1 3 4 2

67. Assertion :- COP of a refrigerator can't be greater

than unity. [AIIMS 2018]

Reason :- Some work is done by working substance.

(1) A (2) B (3) C (4) D

68. Assertion :- In phase transition from liquid to

vapour, potential energy increases. [AIIMS 2018]

Reason :- Temperature is constant during phase

transition.

(1) A (2) B (3) C (4) D

s.n. contents p ge therma ysics

Documents